Kinematics: Question 2

Worked solution

Setting up the problem

Take the direction of the car’s motion as positive throughout.

Part (a): Acceleration during the first $8.0\text{ s}$

The car starts from rest, so $u = 0$, and reaches $v = 24\text{ m s}^{-1}$ after $t = 8.0\text{ s}$. Using $v = u + at$ rearrange for $a$: $a = \frac{v-u}{t} = \frac{24 - 0}{8.0}$ $a = 3.0\text{ m s}^{-2}$

Part (b): Distance travelled during the first $8.0\text{ s}$

Since the acceleration is uniform, the average velocity over this interval is $\tfrac{1}{2}(u+v)$: $s = \left(\frac{u+v}{2}\right)t = \left(\frac{0+24}{2}\right)(8.0)$ $s = 12 \times 8.0 = 96\text{ m}$

As a check, using $s = ut + \tfrac{1}{2}at^2$: $s = (0)(8.0) + \tfrac{1}{2}(3.0)(8.0)^2 = 0 + \tfrac{1}{2}(3.0)(64) = 96\text{ m}$ Both methods agree, so $s = 96\text{ m}$.

Part (c): Deceleration while braking

Now $u = 24\text{ m s}^{-1}$ (the constant speed before braking), $v = 0$ (comes to rest), and $s = 60\text{ m}$. Using $v^2 = u^2 + 2as$ substitute the values: $0 = (24)^2 + 2a(60)$ $0 = 576 + 120a$ $120a = -576$ $a = -4.8\text{ m s}^{-2}$

The negative sign shows that this acceleration acts opposite to the direction of travel, i.e. it is a deceleration. The magnitude of the deceleration is: $4.8\text{ m s}^{-2}$

Final answers

• (a) Acceleration $= \boxed{3.0}\text{ m s}^{-2}$
• (b) Distance travelled $= \boxed{96}\text{ m}$
• (c) Magnitude of deceleration $= \boxed{4.8}\text{ m s}^{-2}$