A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

Question

kap26

1 month ago

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A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

motorcycle’s acceleration? How long does it take to come to a stop

Physics

1 answer:

ValentinkaMS [3.4K] · Answer 1 · 2026-03-02T06:45:49+03:00

Since the motorcycle was at a speed of 36 m/s prior to braking, that marks the initial velocity.
$u = 36 {ms}^{ - 1}$
The motorcycle skidded 23m before coming to a full stop, indicating that
$s = 23m$
As it has ceased motion, the final velocity is zero.

$v = 0{ms}^{ - 1}$
We can apply the 'suvat' formula relevant to linear motion.

${v}^{2} = {u}^{2} + 2as$
Substituting the aforementioned values allows us to find,

${0}^{2} = {36}^{2} + 2a(23)$

$0 = 1296+ 46a$
$46a = - 1296$
$a = - 28.2 {ms}^{ - 2}$

We can apply the formula
$v = u + at$
to calculate the time required for the motorcycle to stop.

$0 = 36 + - 28.2t$
$- 36 = - 28.2t$
$t = 1.3s$