Answer: To calculate the work done by a force, we can use the formula:
Work (W) = Force (F) * Distance (d) * cos(θ)
Where:
Force (F) = 20 N
Distance (d) = 5 m
θ = angle between the direction of the force and the direction of displacement (assuming they are aligned, so cos(θ) = 1)
Enterring in the given values, we have:
W = 20 * 5 * 1
Simplifying the equation, we get:
W = 100 Joules
Therefore, the work done by a force of 20 N in moving an object through a distance of 5 m is 100 Joules.
2. An object is moving with a velocity of 10 m/s. What is its momentum if its mass is 5 kg?
Answer: To calculate the momentum of an object, we can use the formula:
Momentum (p) = Mass (m) * Velocity (v)
Given: Mass of the object (m) = 5 kg Velocity of the object (v) = 10 m/s
Entering in the given values, we have:
p = 5 * 10
Simplifying the equation, we get:
p = 50 kg·m/s
Therefore, the momentum of the object, with a mass of 5 kg and a velocity of 10 m/s, is 50 kg·m/s.
3. A ball is thrown vertically upwards with a velocity of 20 m/s. What is its maximum height?
Answer: To find the maximum height reached by the ball when thrown vertically upwards with a velocity of 20 m/s, we can use the equations of motion.
When the ball reaches its maximum height, its final velocity will be zero. We can use the equation:
v = u + gt
Where:
Enterring in the given values, we have:
W = 20 * 5 * 1
Simplifying the equation, we get:
W = 100 Joules
Therefore, the work done by a force of 20 N in moving an object through a distance of 5 m is 100 Joules.
2. An object is moving with a velocity of 10 m/s. What is its momentum if its mass is 5 kg?
Answer: To calculate the momentum of an object, we can use the formula:
Momentum (p) = Mass (m) * Velocity (v)
Given: Mass of the object (m) = 5 kg Velocity of the object (v) = 10 m/s
Entering in the given values, we have:
p = 5 * 10
Simplifying the equation, we get:
p = 50 kg·m/s
Therefore, the momentum of the object, with a mass of 5 kg and a velocity of 10 m/s, is 50 kg·m/s.
3. A ball is thrown vertically upwards with a velocity of 20 m/s. What is its maximum height?
Answer: To find the maximum height reached by the ball when thrown vertically upwards with a velocity of 20 m/s, we can use the equations of motion.
When the ball reaches its maximum height, its final velocity will be zero. We can use the equation:
v = u + gt
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (20 m/s)
g = acceleration due to gravity (-9.8 m/s², taking negative since it acts in the opposite direction to the initial velocity)
t = time taken to reach the maximum height
Plugging in the values, we have:
0 = 20 - 9.8t
Simplifying the equation, we get:
9.8t = 20
Dividing both sides by 9.8, we find:
t = 2.04 seconds
Now, we can calculate the maximum height (H) using the equation:
H = ut + (1/2)gt²
Plugging in the values, we have:
H = (20)(2.04) + (1/2)(-9.8)(2.04)²
Simplifying the equation, we get:
H = 40.8 - 20.2
H = 20.6 meters
Therefore, the maximum height reached by the ball, when thrown vertically upwards with a velocity of 20 m/s, is approximately 20.6 meter
4. An electric bulb draws a current of 0.5 A when connected to a 220 V supply. What is its power consumption?
Answer: To calculate the power consumption of an electric bulb, we can use the formula:
Power (P) = Current (I) * Voltage (V)
Given:
Plugging in the values, we have:
0 = 20 - 9.8t
Simplifying the equation, we get:
9.8t = 20
Dividing both sides by 9.8, we find:
t = 2.04 seconds
Now, we can calculate the maximum height (H) using the equation:
H = ut + (1/2)gt²
Plugging in the values, we have:
H = (20)(2.04) + (1/2)(-9.8)(2.04)²
Simplifying the equation, we get:
H = 40.8 - 20.2
H = 20.6 meters
Therefore, the maximum height reached by the ball, when thrown vertically upwards with a velocity of 20 m/s, is approximately 20.6 meter
4. An electric bulb draws a current of 0.5 A when connected to a 220 V supply. What is its power consumption?
Answer: To calculate the power consumption of an electric bulb, we can use the formula:
Power (P) = Current (I) * Voltage (V)
Given:
Current (I) = 0.5 A
Entering in the given values, we have:
P = 0.5 * 220
Simplifying the equation, we get:
P = 110 W
Therefore, the power consumption of the electric bulb, when it draws a current of 0.5 A from a 220 V supply, is 110 watts.
5. A convex lens of focal length 20 cm is placed 30 cm from an object. What is the magnification produced by the lens?
Answer: To calculate the magnification produced by a lens, we can use the formula:
Magnification (m) = - (Image distance / Object distance)
Given: Focal length (f) = 20 cm Object distance (u) = 30 cm
We can use the lens formula to find the image distance (v):
1/f = 1/v - 1/u
Plugging in the values, we have:
1/20 = 1/v - 1/30
Simplifying the equation, we get:
1/v = 1/20 + 1/30
1/v = (3 + 2)/60
1/v = 5/60
v = 60/5 = 12 cm
Now, we can calculate the magnification:
m = - (v / u) = - (12 / 30) = - 0.4
Therefore, the magnification produced by the convex lens with a focal length of 20 cm, when the object is placed 30 cm from the lens, is -0.4. The negative sign indicates that the image formed is inverted.
Voltage (V) = 220 V
Entering in the given values, we have:
P = 0.5 * 220
Simplifying the equation, we get:
P = 110 W
Therefore, the power consumption of the electric bulb, when it draws a current of 0.5 A from a 220 V supply, is 110 watts.
5. A convex lens of focal length 20 cm is placed 30 cm from an object. What is the magnification produced by the lens?
Answer: To calculate the magnification produced by a lens, we can use the formula:
Magnification (m) = - (Image distance / Object distance)
Given: Focal length (f) = 20 cm Object distance (u) = 30 cm
We can use the lens formula to find the image distance (v):
1/f = 1/v - 1/u
Plugging in the values, we have:
1/20 = 1/v - 1/30
Simplifying the equation, we get:
1/v = 1/20 + 1/30
1/v = (3 + 2)/60
1/v = 5/60
v = 60/5 = 12 cm
Now, we can calculate the magnification:
m = - (v / u) = - (12 / 30) = - 0.4
Therefore, the magnification produced by the convex lens with a focal length of 20 cm, when the object is placed 30 cm from the lens, is -0.4. The negative sign indicates that the image formed is inverted.
6.What is the frequency of a wave with a wavelength of 2 m and a speed of 300 m/s?
Answer: To find the frequency of a wave, we can use the formula:
Frequency (f) = Speed (v) / Wavelength (λ)
Given: Wavelength (λ) = 2 m Speed (v) = 300 m/s
Plugging in the given values, we have:
f = 300 / 2
Simplifying the equation, we get:
f = 150 Hz
Therefore, the frequency of the wave, with a wavelength of 2 m and a speed of 300 m/s, is 150 Hz.
7. A spring of spring constant 100 N/m is compressed by 0.2 m. What is the potential energy stored in the spring?
Answer: To calculate the potential energy stored in a spring, we can use the formula:
Potential energy (PE) = (1/2) * k * x²
Where: k is the spring constant (100 N/m in this case) x is the displacement from the equilibrium position (0.2 m in this case)
Enter in the given values, we have:
PE = (1/2) * 100 * (0.2)²
Simplifying the equation, we get:
PE = (1/2) * 100 * 0.04
PE = 2 joules
Therefore, the potential energy stored in the spring, with a spring constant of 100 N/m and compressed by 0.2 m, is 2 joules.
8. A car is moving with a speed of 10 m/s. What is its kinetic energy if its mass is 1000 kg?
Answer: To calculate the kinetic energy of the car, we can use the formula:
Kinetic energy (KE) = 1/2 * mass * velocity^2
Given: Mass of the car (m) = 1000 kg Velocity of the car (v) = 10 m/s
Plugging in the values, we have:
KE = 1/2 * 1000 * (10)^2
Simplifying the equation, we get:
KE = 1/2 * 1000 * 100
KE = 50,000 joules
Therefore, the kinetic energy of the car, with a mass of 1000 kg and a speed of 10 m/s, is 50,000 joules.
9. A body is moving along a straight line. Its initial velocity is 10 m/s and it accelerates at 2 m/s2. What is its velocity after 5 seconds?
Answer: To find the velocity of the body after 5 seconds of acceleration, we can use the equation of motion:
v = u + at
Where:
Answer: To find the frequency of a wave, we can use the formula:
Frequency (f) = Speed (v) / Wavelength (λ)
Given: Wavelength (λ) = 2 m Speed (v) = 300 m/s
Plugging in the given values, we have:
f = 300 / 2
Simplifying the equation, we get:
f = 150 Hz
Therefore, the frequency of the wave, with a wavelength of 2 m and a speed of 300 m/s, is 150 Hz.
7. A spring of spring constant 100 N/m is compressed by 0.2 m. What is the potential energy stored in the spring?
Answer: To calculate the potential energy stored in a spring, we can use the formula:
Potential energy (PE) = (1/2) * k * x²
Where: k is the spring constant (100 N/m in this case) x is the displacement from the equilibrium position (0.2 m in this case)
Enter in the given values, we have:
PE = (1/2) * 100 * (0.2)²
Simplifying the equation, we get:
PE = (1/2) * 100 * 0.04
PE = 2 joules
Therefore, the potential energy stored in the spring, with a spring constant of 100 N/m and compressed by 0.2 m, is 2 joules.
8. A car is moving with a speed of 10 m/s. What is its kinetic energy if its mass is 1000 kg?
Answer: To calculate the kinetic energy of the car, we can use the formula:
Kinetic energy (KE) = 1/2 * mass * velocity^2
Given: Mass of the car (m) = 1000 kg Velocity of the car (v) = 10 m/s
Plugging in the values, we have:
KE = 1/2 * 1000 * (10)^2
Simplifying the equation, we get:
KE = 1/2 * 1000 * 100
KE = 50,000 joules
Therefore, the kinetic energy of the car, with a mass of 1000 kg and a speed of 10 m/s, is 50,000 joules.
9. A body is moving along a straight line. Its initial velocity is 10 m/s and it accelerates at 2 m/s2. What is its velocity after 5 seconds?
Answer: To find the velocity of the body after 5 seconds of acceleration, we can use the equation of motion:
v = u + at
Where:
v = final velocity (to be determined)
u = initial velocity (10 m/s)
a = acceleration (2 m/s²)
t = time (5 seconds)
Enter in the given values, we have:
v = 10 + (2)(5)
Simplifying the equation, we get:
v = 10 + 10
v = 20 m/s
Therefore, the velocity of the body after 5 seconds, starting with an initial velocity of 10 m/s and accelerating at 2 m/s², is 20 m/s.
10. A car is moving with a speed of 20 m/s. It accelerates uniformly at 4 m/s2 for 10 seconds. What is its final velocity?
Answer: To find the final velocity of the car after accelerating uniformly at 4 m/s² for 10 seconds, we can use the equation of motion:
v = u + at
Where:
Enter in the given values, we have:
v = 10 + (2)(5)
Simplifying the equation, we get:
v = 10 + 10
v = 20 m/s
Therefore, the velocity of the body after 5 seconds, starting with an initial velocity of 10 m/s and accelerating at 2 m/s², is 20 m/s.
10. A car is moving with a speed of 20 m/s. It accelerates uniformly at 4 m/s2 for 10 seconds. What is its final velocity?
Answer: To find the final velocity of the car after accelerating uniformly at 4 m/s² for 10 seconds, we can use the equation of motion:
v = u + at
Where:
v = final velocity (to be determined)
u = initial velocity (20 m/s)
a = acceleration (4 m/s²)
t = time (10 seconds)
Enter in the given values, we have:
v = 20 + (4)(10)
Simplifying the equation, we get:
v = 20 + 40
v = 60 m/s
Therefore, the final velocity of the car, after accelerating uniformly at 4 m/s² for 10 seconds from an initial speed of 20 m/s, is 60 m/s.
11. A particle moves in a straight line with an initial velocity of 10 m/s. It experiences a constant acceleration of 5 m/s2. Find the distance travelled by the particle in 6 seconds.
Answer: To find the distance traveled by the particle in 6 seconds, we can use the equation of motion:
s = ut + (1/2)at²
Where:
Enter in the given values, we have:
v = 20 + (4)(10)
Simplifying the equation, we get:
v = 20 + 40
v = 60 m/s
Therefore, the final velocity of the car, after accelerating uniformly at 4 m/s² for 10 seconds from an initial speed of 20 m/s, is 60 m/s.
11. A particle moves in a straight line with an initial velocity of 10 m/s. It experiences a constant acceleration of 5 m/s2. Find the distance travelled by the particle in 6 seconds.
Answer: To find the distance traveled by the particle in 6 seconds, we can use the equation of motion:
s = ut + (1/2)at²
Where:
s = distance travelled
u = initial velocity (10 m/s)
a = acceleration (5 m/s²)
t = time (6 seconds)
Enter in the given values, we have:
s = (10)(6) + (1/2)(5)(6)²
Simplifying the equation, we get:
s = 60 + (1/2)(5)(36)
s = 60 + 90
s = 150 meters
Therefore, the particle will travel a distance of 150 meters in 6 seconds, given that it has an initial velocity of 10 m/s and experiences a constant acceleration of 5 m/s².
12. A stone is thrown vertically upwards with an initial velocity of 20 m/s. Find the maximum height attained by the stone.
Answer: To find the maximum height attained by the stone when thrown vertically upwards with an initial velocity of 20 m/s, we can use the equations of motion.
When the stone reaches its maximum height, its final velocity will be zero. We can use the equation:
v = u + gt
Where:
Enter in the given values, we have:
s = (10)(6) + (1/2)(5)(6)²
Simplifying the equation, we get:
s = 60 + (1/2)(5)(36)
s = 60 + 90
s = 150 meters
Therefore, the particle will travel a distance of 150 meters in 6 seconds, given that it has an initial velocity of 10 m/s and experiences a constant acceleration of 5 m/s².
12. A stone is thrown vertically upwards with an initial velocity of 20 m/s. Find the maximum height attained by the stone.
Answer: To find the maximum height attained by the stone when thrown vertically upwards with an initial velocity of 20 m/s, we can use the equations of motion.
When the stone reaches its maximum height, its final velocity will be zero. We can use the equation:
v = u + gt
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (20 m/s)
g = acceleration due to gravity (-9.8 m/s², taking negative since it acts in the opposite direction to the initial velocity)
t = time taken to reach the maximum height
Plugging in the values, we have:
0 = 20 - 9.8t
Simplifying the equation, we get:
9.8t = 20
Dividing both sides by 9.8, we find:
t = 2.04 seconds
Now, we can calculate the maximum height (H) using the equation:
H = ut + (1/2)gt²
Plugging in the values, we have:
H = (20)(2.04) + (1/2)(-9.8)(2.04)²
Simplifying the equation, we get:
H = 20.4 - 20.2
H = 0.2 meters
Therefore, the maximum height attained by the stone, when thrown vertically upwards with an initial velocity of 20 m/s, is approximately 0.2 meters.
13. A ball is dropped from a height of 20 meters. Find the time taken by the ball to hit the ground.
Answer: To find the time taken by the ball to hit the ground when dropped from a height of 20 meters, we can use the equation of motion:
s = ut + 1/2gt²
Where:
Plugging in the values, we have:
0 = 20 - 9.8t
Simplifying the equation, we get:
9.8t = 20
Dividing both sides by 9.8, we find:
t = 2.04 seconds
Now, we can calculate the maximum height (H) using the equation:
H = ut + (1/2)gt²
Plugging in the values, we have:
H = (20)(2.04) + (1/2)(-9.8)(2.04)²
Simplifying the equation, we get:
H = 20.4 - 20.2
H = 0.2 meters
Therefore, the maximum height attained by the stone, when thrown vertically upwards with an initial velocity of 20 m/s, is approximately 0.2 meters.
13. A ball is dropped from a height of 20 meters. Find the time taken by the ball to hit the ground.
Answer: To find the time taken by the ball to hit the ground when dropped from a height of 20 meters, we can use the equation of motion:
s = ut + 1/2gt²
Where:
s = vertical displacement (20 meters)
u = initial velocity (0 m/s, as the ball is initially at rest when dropped)
g = acceleration due to gravity (approximately -9.8 m/s², negative since it acts in the opposite direction to the displacement)
t = time taken to hit the ground
enter in the given values, we have:
20 = 0 * t + 1/2 * (-9.8) * t²
Simplifying the equation, we get:
20 = -4.9t²
Dividing both sides by -4.9, we find:
t² = -20 / -4.9
t² = 4.08
Taking the square root of both sides, we get:
t = √4.08
t = 2.02 seconds
Therefore, it will take approximately 2.02 seconds for the ball to hit the ground when dropped from a height of 20 meters.
enter in the given values, we have:
20 = 0 * t + 1/2 * (-9.8) * t²
Simplifying the equation, we get:
20 = -4.9t²
Dividing both sides by -4.9, we find:
t² = -20 / -4.9
t² = 4.08
Taking the square root of both sides, we get:
t = √4.08
t = 2.02 seconds
Therefore, it will take approximately 2.02 seconds for the ball to hit the ground when dropped from a height of 20 meters.
14. A train is moving with a speed of 72 km/h. If it accelerates at 2 m/s2, how much time will it take to attain a speed of 108 km/h?
Answer: To find the time it takes for the train to attain a speed of 108 km/h, we can convert the speeds from km/h to m/s and then use the equation of kinematics:
v = u + at
Where:
Answer: To find the time it takes for the train to attain a speed of 108 km/h, we can convert the speeds from km/h to m/s and then use the equation of kinematics:
v = u + at
Where:
v = final velocity (108 km/h converted to m/s)
u = initial velocity (72 km/h converted to m/s)
a = acceleration (2 m/s²)
t = time
Converting the velocities: 108 km/h = (108 * 1000) / 3600 m/s = 30 m/s 72 km/h = (72 * 1000) / 3600 m/s = 20 m/s
Now, we can substitute the values into the equation and solve for time:
30 = 20 + 2t
Simplifying the equation, we get:
2t = 30 - 20
2t = 10
Dividing both sides by 2, we find:
t = 5 seconds
Therefore, it will take approximately 5 seconds for the train to attain a speed of 108 km/h, given that it initially moves at a speed of 72 km/h and accelerates at a rate of 2 m/s².
Converting the velocities: 108 km/h = (108 * 1000) / 3600 m/s = 30 m/s 72 km/h = (72 * 1000) / 3600 m/s = 20 m/s
Now, we can substitute the values into the equation and solve for time:
30 = 20 + 2t
Simplifying the equation, we get:
2t = 30 - 20
2t = 10
Dividing both sides by 2, we find:
t = 5 seconds
Therefore, it will take approximately 5 seconds for the train to attain a speed of 108 km/h, given that it initially moves at a speed of 72 km/h and accelerates at a rate of 2 m/s².
15. A car is moving with a speed of 25 m/s. It comes to rest after moving a distance of 50 m. What is the acceleration of the car?
Answer: To find the acceleration of the car, we can use the equation of kinematics:
v^2 = u^2 + 2as
Where:
Answer: To find the acceleration of the car, we can use the equation of kinematics:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (25 m/s)
a = acceleration (to be determined)
s = distance travelled (50 m)
entering in the given values, we have:
0 = (25)^2 + 2a(50)
Simplifying the equation, we get:
0 = 625 + 100a
Rearranging the equation, we find:
100a = -625
Dividing both sides by 100, we get:
a = -6.25 m/s^2
Therefore, the acceleration of the car, as it comes to rest after moving a distance of 50 m with an initial speed of 25 m/s, is approximately -6.25 m/s^2. The negative sign indicates that the car is decelerating (slowing down)
entering in the given values, we have:
0 = (25)^2 + 2a(50)
Simplifying the equation, we get:
0 = 625 + 100a
Rearranging the equation, we find:
100a = -625
Dividing both sides by 100, we get:
a = -6.25 m/s^2
Therefore, the acceleration of the car, as it comes to rest after moving a distance of 50 m with an initial speed of 25 m/s, is approximately -6.25 m/s^2. The negative sign indicates that the car is decelerating (slowing down)
16. A ball is thrown horizontally with a speed of 20 m/s. If it travels a horizontal distance of 40 meters before hitting the ground, what is the time of flight of the ball?
Answer: To find the time of flight of the ball, we can use the equation for horizontal motion:
d = v * t
Where:
Answer: To find the time of flight of the ball, we can use the equation for horizontal motion:
d = v * t
Where:
d = horizontal distance traveled (40 meters)
v = horizontal velocity (20 m/s)
t = time of flight
Plugging in the given values, we have:
40 = 20 * t
Simplifying the equation, we get:
t = 40 / 20
t = 2 seconds
Therefore, the time of flight of the ball, when thrown horizontally with a speed of 20 m/s and travelling a horizontal distance of 40 meters, is 2 seconds.
17. A particle moves in a straight line with an initial velocity of 5 m/s. If it experiences a constant acceleration of 10 m/s2, what is its velocity after 4 seconds?
Answer: To find the velocity of the particle after 4 seconds, we can use the equation of kinematics:
v = u + at
Where:
Plugging in the given values, we have:
40 = 20 * t
Simplifying the equation, we get:
t = 40 / 20
t = 2 seconds
Therefore, the time of flight of the ball, when thrown horizontally with a speed of 20 m/s and travelling a horizontal distance of 40 meters, is 2 seconds.
17. A particle moves in a straight line with an initial velocity of 5 m/s. If it experiences a constant acceleration of 10 m/s2, what is its velocity after 4 seconds?
Answer: To find the velocity of the particle after 4 seconds, we can use the equation of kinematics:
v = u + at
Where:
v = final velocity (to be determined)
u = initial velocity (5 m/s)
a = acceleration (10 m/s²)
t = time (4 seconds)
Plugging in the values, we have:
v = 5 + (10)(4)
Simplifying the equation, we get:
v = 5 + 40
v = 45 m/s
Therefore, the velocity of the particle after 4 seconds, when experiencing a constant acceleration of 10 m/s² with an initial velocity of 5 m/s, is 45 m/s.
18. A stone is thrown vertically upwards with an initial velocity of 15 m/s. Find the time taken by the stone to reach the highest point.
Answer:To find the time taken by the stone to reach the highest point, we can use the equation of kinematics:
v = u + at
Where:
Plugging in the values, we have:
v = 5 + (10)(4)
Simplifying the equation, we get:
v = 5 + 40
v = 45 m/s
Therefore, the velocity of the particle after 4 seconds, when experiencing a constant acceleration of 10 m/s² with an initial velocity of 5 m/s, is 45 m/s.
18. A stone is thrown vertically upwards with an initial velocity of 15 m/s. Find the time taken by the stone to reach the highest point.
Answer:To find the time taken by the stone to reach the highest point, we can use the equation of kinematics:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (15 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s²)
t = time
At the highest point, the stone momentarily comes to rest, so its final velocity is 0 m/s. Plugging in the values, we have:
0 = 15 + (-9.8)t
Simplifying the equation, we get:
-15 = -9.8t
Dividing both sides by -9.8, we find:
t = 1.53 seconds
Therefore, it will take approximately 1.53 seconds for the stone to reach the highest point when thrown vertically upwards with an initial velocity of 15 m/s.
19. A car is moving with a velocity of 15 m/s. If it decelerates at 2 m/s2, how much time will it take to come to rest?
Answer: To determine the time it takes for the car to come to rest, we can use the equation of kinematics:
v = u + at
Where:
At the highest point, the stone momentarily comes to rest, so its final velocity is 0 m/s. Plugging in the values, we have:
0 = 15 + (-9.8)t
Simplifying the equation, we get:
-15 = -9.8t
Dividing both sides by -9.8, we find:
t = 1.53 seconds
Therefore, it will take approximately 1.53 seconds for the stone to reach the highest point when thrown vertically upwards with an initial velocity of 15 m/s.
19. A car is moving with a velocity of 15 m/s. If it decelerates at 2 m/s2, how much time will it take to come to rest?
Answer: To determine the time it takes for the car to come to rest, we can use the equation of kinematics:
v = u + at
Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (15 m/s)
a = acceleration (deceleration in this case, -2 m/s²)
t = time
Rearranging the equation to solve for time (t), we have:
0 = 15 + (-2)t
Simplifying the equation, we get:
-15 = -2t
Dividing both sides by -2, we find:
t = 7.5 seconds
Therefore, it will take approximately 7.5 seconds for the car to come to rest when decelerating at 2 m/s² from an initial velocity of 15 m/s.
Rearranging the equation to solve for time (t), we have:
0 = 15 + (-2)t
Simplifying the equation, we get:
-15 = -2t
Dividing both sides by -2, we find:
t = 7.5 seconds
Therefore, it will take approximately 7.5 seconds for the car to come to rest when decelerating at 2 m/s² from an initial velocity of 15 m/s.
20. A motorcyclist accelerates from rest at a rate of 2 m/s2. After 5 seconds, what is its speed?
Answer: To find the speed of the motorcyclist after 5 seconds of acceleration, we can use the equation of motion:
Speed (v) = Initial velocity (u) + Acceleration (a) * Time (t)
Given: Initial velocity (u) = 0 m/s (rest) Acceleration (a) = 2 m/s² Time (t) = 5 seconds
Entering in the given values, we have:
v = 0 + 2 * 5
Simplifying the equation, we get:
v = 10 m/s
Therefore, the speed of the motorcyclist after 5 seconds of accelerating from rest at a rate of 2 m/s² is 10 m/s.
Speed (v) = Initial velocity (u) + Acceleration (a) * Time (t)
Given: Initial velocity (u) = 0 m/s (rest) Acceleration (a) = 2 m/s² Time (t) = 5 seconds
Entering in the given values, we have:
v = 0 + 2 * 5
Simplifying the equation, we get:
v = 10 m/s
Therefore, the speed of the motorcyclist after 5 seconds of accelerating from rest at a rate of 2 m/s² is 10 m/s.
