Kinematics of Motion in a Straight Line: Question 6

Worked solution

Part (a): Finding the deceleration

Take the direction of travel as positive. We know:

• $u = 24\text{ m s}^{-1}$ (initial velocity)
• $v = 0\text{ m s}^{-1}$ (comes to rest)
• $s = 180\text{ m}$ (distance travelled while braking)

Since time is not involved, use $v^2 = u^2 + 2as$:

$0 = 24^2 + 2a(180) = 576 + 360a$

$360a = -576 \implies a = -\frac{576}{360} = -1.6$

So the train’s acceleration is $-1.6\text{ m s}^{-2}$, i.e. a deceleration of $1.6\text{ m s}^{-2}$.

Recompute independently as a check: $360 \times 1.6 = 360 \times 1 + 360\times0.6 = 360+216=576$, which matches the right-hand side exactly, confirming $a=-1.6\text{ m s}^{-2}$.

Part (b): Finding the time to come to rest

Using $v = u + at$ with $u=24$, $v=0$, $a=-1.6$:

$0 = 24 + (-1.6)t$

$1.6t = 24 \implies t = \frac{24}{1.6} = 15$

So the train takes $15$ s to come to rest.

Recompute independently as a check: $1.6 \times 15 = 1.6\times10 + 1.6\times5 = 16+8=24$, which matches $u=24$ exactly, confirming $t=15$ s.

Part (c): Average velocity check

For constant acceleration, the average velocity is the mean of the initial and final velocities:

$\bar v = \frac{u+v}{2} = \frac{24+0}{2} = 12\text{ m s}^{-1}$

The distance travelled is (average velocity) $\times$ (time taken):

$s = \bar v \times t = 12 \times 15 = 180$

This equals the $180$ m given in the question, confirming that $t=15$ s from part (b) (and $a=-1.6\text{ m s}^{-2}$ from part (a)) are correct.

Final answers

• (a) Deceleration $=\boxed{1.6\text{ m s}^{-2}}$
• (b) Time to come to rest $=\boxed{15\text{ s}}$
• (c) Average velocity $=\boxed{12\text{ m s}^{-1}}$; $12\times15=180$ m, matching the given distance