Question

a person ride a motorbike at the speed of 30 metre per second the person applies a brake and the velocity of motorbike comes down 20 metre per second in 3 seconds what is the magnitude of acceleration of motorbike​

Answer 1

Answer:

☞Acceleration = -0.333 m/$s^2$

Given:

☞Initial Velocity (u) = 30 m/s

☞Final Velocity (v) = 20 m/s

☞Time taken for change in velocity (t) = 3 seconds

Concept:

We know that , acceleration is the rate of change of velocity.

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

Explanation:

Using the above concept and substituting the given values, we get:

a = $\dfrac{20-30}{3}$ m/$s^2$

a = $\dfrac{-10}{3}$ m/$s^2$

a = -0.333 m/$s^2$

acceleration = -0.333 m/$s^2$

Therefore, the answer is -0.333 m/$s^2$

Other Formulas:

Equations Of Motion:

1. v = u + at
2. $s = ut + \dfrac{1}{2} {at}^{2}$
3. ${v}^{2} - {u}^{2} = 2as$

where

u = initial velocity of the body

v = final velocity of the body

a = acceleration of the body

t = time taken for change

s = distance covered

g = acceleration due to gravity

In some cases , if acceleration is not given and acceleration due to gravity is given , then we substitute the value of g.

v = u + gt

$s = ut + \dfrac{1}{2} {gt}^{2}$

${v}^{2} - {u}^{2} = 2gh$

In laws of equation, if the body is going up against acceleration due to gravity, then:

v = u - gt

$s = ut - \dfrac{1}{2} {gt}^{2}$

${v}^{2} - {u}^{2} = -2gh$

Answer 2

Answer:

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