**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 traveled by the 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 = ?

$$\begin{array}{l}s=ut+\frac{1}{2}a{t}^{2}\\ s=(0)(10)+\frac{1}{2}(3){\left(10\right)}^{2}=150m\end{array}$$

__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.

$$\begin{array}{l}v=u+at\\ v=(0)+(3)(10)=30m{s}^{-1}\end{array}$$ 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 = ?

$$\begin{array}{l}s=\frac{1}{2}(u+v)t\\ s=\frac{1}{2}(30+0)(5)=75m\end{array}$$Total distance travelled = 150 + 150 + 75 = 375m

### Reference

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