Question

A motor cyclist travelling at 12m/s decelerates at 3m/s² a) How long does it take to come to rest?b) How far does he travel in coming to rest?

50 points♥♥ wrong answer reported

Answer 1

Given:-

• Initial velocity (u) = 12 m/s
• Acceleration (a) = - 3 m/s² {Negative acceleration}

To Find: Time it will take to become stationary; distance it will cover at that time.

(A) :-

Final velocity (v) = 0 m/s

We know,

v = u + at

or, t = - u/a

∴ t = (- 12 m/s)/(- 3 m/s²)

→ t = 4 s

So, the particle would take 4 seconds to stop.

(B) :-

We know,

s = ut + ½at²

∴ s = (12 m/s)(4 s) + ½(- 3 m/s²)(4 s)²

→ s = 48 m + ½(- 3m/s²)(16 s²)

→ s = 48 m + (- 3 m/s²)(8 s²)

→ s = 48 m - 24 m

→ s = 24 m

So, it will cover 24 metres.

Answer 2

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

Initial velocity = 12m/s

Final velocity = zero (rest)

Acceleration = -3m/s²

[Negative sign shows retardation]

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

⛥ Time taken by motor cyclist to come at rest position.

⛥ Distance travelled by motor cyclist before it is brought to rest.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

ATQ, Acceleration is said to be constant throughout the motion, So we can easily apply equation of kinematics to solve this question.

◈ Calculation of Time :

$:\implies\tt\:v=u+at$

$:\implies\tt\:0=12+(-3)t$

$:\implies\boxed{\bf{t=4\:s}}$

◈ Calculation of Distance :

$:\implies\tt\:v^2-u^2=2as$

$:\implies\tt\:0^2-12^2=2(-3)s$

$:\implies\tt\:144=6s$

$:\implies\boxed{\bf{s=24\:m}}$