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?  

Answer:
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 = ?
\[\begin{array}{l}
s = \frac{1}{2}(u + v)t\\
s = \frac{1}{2}(30 + 0)(5) = 75m
\end{array}\]