A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?

t = 9.14 s

Explanation:

We first analyze the accelerating motion by applying first equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car before turning off engine

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 25 m/s²

t₁ = time taken in accelerating motion

Therefore,

Vf₁ = 25t₁   equation (1)

Now, we apply second equation of motion:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = distance covered during accelerating motion

Therefore,

s₁ = (0)t₁ + (1/2)(25)t₁²

s₁ = 12.5 t₁²   equation (2)

Now, we analyze the decelerating motion by applying first equation of motion:

Vf₂ = Vi₂ + a₂t₂

where,

Vf₂ = Final Speed of Car = 0 m/s

Vi₂ = Initial Speed of Car after turning off engine

a₂ = deceleration of car = - 3 m/s²

t₂ = time taken in decelerating motion

Therefore,

Vi₂ = 3t₂   equation (3)

Now, we apply second equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered during decelerating motion

Therefore,

s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²

s₂ = Vi₂ t₂ - 1.5 t₂²  

using equation (3):

s₂ = 3 t₂² - 1.5 t₂²

s₂ = 1.5 t₂²   equation (4)

Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):

Vf₁ = Vi₂

using equation (1) and equation (3):

25 t₁ = 3 t₂

t₁ = 0.12 t₂   equation (5)

Also, we know that sum of the distances is 200 m:

s₁ + s₂ = 200

using equation (2) and equation (4):

12.5 t₁² + 1.5 t₂² = 200

using equation (5):

12.5 (0.12 t₂²) + 1.5 t₂² = 200

3 t₂² = 200

t₂² = 200/3

t₂ = 8.16 s

substitute this in equation (5):

t₁ = 0.12(8.16 s)

t₁ = 0.97 s

Hence, the minimum time required for this motion is:

t = t₁ + t₂ = 0.97 s + 8.16 s

t = 9.14 s

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )

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