Question

2. A car takes 8 s to slow down from a speed of 15 ms¹ to 5 ms¹ as it approaches a

junction.
a) What is the acceleration of the car?
b) How far does the car travel while it is slowing down?

Answer 1

Solution below.

Step-by-step explanation:

First, lets list down all the kinematics equations.

`v = u + at \\ {v}^(2) = {u}^(2) + 2as \\ s = ut + (1)/(2) a {t}^(2)`

where:

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement (or distance(

t is the time taken

We can use any of the equation based on the given variables in the question.

a) Given t = 8 seconds, v = 5 m/s and u = 15 m/s, we can use the first equation to solve for a, the acceleration.

v = u + at

5 = 15 + a × 8

a × 8 + 15 = 5

a × 8 = 5 - 15

a × 8 = - 10

a = -10 ÷ 8

a =

`- 1.25 \frac{m}{ {s}^(2) }`

(Deccelarating)

b) Now, we can use the acceleration found in part a and the 3rd equation to solve for s, the distance.

`s = ut + (1)/(2) a {t}^(2) \\ s = (15)(8) + (1)/(2) ( - 1.25) {(8)}^(2) \\ = 80m`