Question

A cyclist riding at a speed of 5m/s braked with uniform deceleration and stopped in 3m. How long did she take to stop?

Answer 1

Answer:

According to third equation of motion

v^2-u^2 = 2aS

0^2 - 5^2 = 2aS

0 - 25 = (2×3) S

-25 = 6 S

-25/6 = S

-41.6

Answer 2

Concept:

First and second kinematic equations are applied. The equations are- v = u + at and v² = u² + 2as

Given:

velocity, u = 5m/s

displacement = 3m

Find:

We need to determine the time taken by the cyclist to stop.

Solution:

The motion of an object with constant acceleration is described by a set of equations known as kinematic equations.

The second kinematic equation is expressed as, v² = u² + 2as

It can be rearranged as, a =  v² - u²/ 2s

a = 0 - (5)² / 2(3)

a = -25 / 6

a = -4.16m/s²

From the value of acceleration, it is possible to determine the time taken by the cyclist to stop

v = u + at

0 = 5 - (4.16) t

4.16 t = 5

t = 5/ 4.16

t = 1.2 secs

Thus, the time taken by the cyclist to stop is 1.2 secs.

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