 # Motion with Uniform Acceleration – Challenging Question 1

Challenging Question 1:
A car starts from rest and accelerates at a constant acceleration of 3 ms-2 for 10 seconds. The car then travels at a constant velocity for 5 seconds. The brakes are then applied and the car stops in 5 seconds. What is the total distance travelled by car?

It’s advisable to list down all the information that we have.

In this question, there are 3 stages of motion.

0s – 10s
Initial velocity, u = 0 (Because the car start from rest)
Acceleration, a = 3 ms-2
Time taken, t = 10s
Displacement, s = ?

$\begin{array}{l} s = ut + \frac{1}{2}a{t^2}\\ s = \left( 0 \right)\left( {10} \right) + \frac{1}{2}\left( 3 \right){\left( {10} \right)^2} = 150m{\rm{ }} \end{array}$
10s – 15s
In this stage, the car moves with a constant velocity. The velocity is equal to the final velocity of previous stage. We can use the equation (v = u + at) to determine the velocity.
$\begin{array}{l} v = u + at\\ v = (0) + (3)(10) = 30m{s^{ – 1}}{\rm{ }} \end{array}$
Therefore,
$\begin{array}{l} v = \frac{s}{t}\\ s = vt = (30)(5) = 150m \end{array}$
15s – 20s
In this stage, the car undergoes deceleration.
Initial velocity, u = 30 ms-1
Final velocity, v = 0 (The car stop at the end)
Time taken, t = 5s
Displacement, s = ?
$\begin{array}{l} s = \frac{1}{2}(u + v)t\\ s = \frac{1}{2}(30 + 0)(5) = 75m \end{array}$

### Reference

$\begin{array}{l} v = u + at{\rm{ }}\\ s = ut + \frac{1}{2}a{t^2}\\ {v^2} = {u^2} + 2as\\ s = \frac{1}{2}(u + v)t \end{array}$