# 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 traveled by the 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 = ?

$s=ut+ 1 2 a t 2 s=(0)(10)+ 1 2 (3) ( 10 ) 2 =150m$
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.
$v=u+at v=(0)+(3)(10)=30m s −1$ Therefore, $v= s t s=vt=(30)(5)=150m$
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 = ?
Total distance travelled = 150 + 150 + 75 = 375m

### Reference

$v=u+at s=ut+ 1 2 a t 2 v 2 = u 2 +2as s= 1 2 (u+v)t$