Physics, asked by mkpanda12, 3 months ago

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.​

Answers

Answered by snehitha2
7

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.

Answered by NewGeneEinstein
3
  • 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

Similar questions