Question

A motorcycle was initially moving with a velocity of 80 m/s, suddenly after seeing a child in front of him, the motorcycle rider applied a brake and motorcycle comes in rest after 16 seconds. Calculate (a) acceleration of the motorcycle (b) distance covered.​

Answer 1

Answer:

(a) -5 m/s²

(b) 640 m

Explanation:

Given :

A motorcycle was initially moving with a velocity of 80 m/s, suddenly after seeing a child in front of him, the motorcycle rider applied a brake and motorcycle comes in rest after 16 seconds.

To find :

(a) acceleration of the motorcycle

(b) distance covered

Solution :

Initial velocity, u = 80 m/s

final velocity, v = 0 m/s

time taken, t = 16 seconds

Using v = u + at :

0 = 80 + a(16)

16a = -80

  a = -80/16

  a = -5 m/s²

Therefore, the acceleration of the motorcycle is -5 m/s²

Now applying v² - u² = 2as :

0² - 80² = 2(-5)(s)

-6400 = -10s

  10s = 6400

  s = 6400/10

  s = 640 m

∴ The distance covered is 640 m.

Answer 2

• initial velocity=u=80m/s
• Final velocity=v=0m/s
• Time=t=16s
• Acceleration=a=?
• Distance=s=?

We know

$\boxed{\sf a=\dfrac{v-u}{t}}$

$\\ \sf\longmapsto a=\dfrac{0-80}{16}$

$\\ \sf\longmapsto a=\dfrac{-80}{16}$

$\\ \sf\longmapsto a=-5m/s^2$

Now using 2nd equation of kinematics

$\boxed{\sf s=ut+\dfrac{1}{2}at^2}$

$\\ \sf\longmapsto s=80(16)+\dfrac{1}{2}\times (-5)(16)^2$

$\\ \sf\longmapsto s=1280+(-5)(128)$

$\\ \sf\longmapsto s=1280+(-640)$

$\\ \sf\longmapsto s=1280-640$

$\\ \sf\longmapsto s=640m$