Physics question on Kinematic equations?

Question

Here's the question:

A car traveling 45 km/h slows down at a constant 0.50 m/s^2 just by "letting up on the gas." Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.

My question is: which of the kinematic equations do I use for a, b and c?

Can someone explain this to me?

Answer 1

Since you are only asking for the equations, here they are:

<< a) the distance the car coasts before it stops, >>

Vf^2 - Vo^2 = 2as

where

Vf = final velocity = 0 (since the car will eventually stop)

Vo = initial velocity = 45 kph (convert to m/sec)

a = acceleration = -0.50 m/sec^2 (negative because it is actually a deceleration, i.e.. the car is slowing down)

s = stopping distance

<< (b) the time it takes to stop, >>

Vf - Vo = aT

OR

T = (Vf - Vo)/a

where

T = time for car to stop

and all the other terms have been previously defined.

<< (c) the distance it travels during the first and fifth seconds. >>

S = VoT + (1/2)(a)T^2

Substitute

T = 1 and T = 5 to get the respective answers.