Question

A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2 . The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s, how fast is the car going at the end of the braking period?

Answer 1

The final velocity of the car is +1.5 m/s.

Explanation:

We have to divide the problem into two parts.

In the first part, the car starts from rest and accelerates for 5.0 s. We can find the final velocity of the car after this first part using the suvat equation:

$v = u +at$

where

v is the final velocity

u = 0 is the initial velocity

$a=+1.5 m/s^2$ is the acceleration

t = 5.0 s is the time

Substituting,

$v=0+(1.5)(5.0)=7.5 m/s$

In the second part, the brakes are applied, so the car decelerates for 3.0 s, and the final velocity is given by

$v' = v + at$

where

v' is the final velocity

v = 7.5 m/s is the velocity at the beginning of this part

$a=-2.0 m/s^2$ is the deceleration

t = 3.0 s is the time

Substituting,

$v'=7.5+(-2.0)(3.0)=+1.5 m/s$

So, the final velocity is +1.5 m/s.

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