Question

A car having an initial speed of 16 meters per second is uniformly brought to rest in 4.0 seconds. How far does the car travel during this 4.0 second interval

Answer 1

let d is the distance covered by the car
Given vi = 16, vf = 0, t = 4 
Byusing the following formula we can calculate the acceleration;
vf - vi = at
a = vf -vi / t
=0 - 16 / 4
a = -4m/s²
Now use this formula to calculate distance
d= 1/2at² + vit
=1/2 (-4) (4)² + 16(4)
=-32 + 64
d = 32 m
So, the car travels 32m during this 4 second interval.

Answer 2

The distance travelled by the car before stopping is 32 m

Given data:

The initial speed of car is, u = 16 m/s.

The final speed of car is, v = 0 m/s. (As car comes to rest finally)

The time interval is, t = 4.0 s.

Apply the first kinematic equation of motion as,

$v=u+at$

Here, a is the acceleration of car.

Solving as,

$0=16+a(4)\\a=-4 \;\rm m/s^{2}$

Now, apply the third kinematic equation of motion as,

$v^{2}=u^{2}+2as \\0^{2}=16^{2}+2(-4)s\\8s=256\\s =32\;\rm m$

Thus, the distance travelled by the car before stopping is 32 m.

Learn more about kinematic equations of motion here:

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