Question

A person traveling at 43•2 km/h applies the brakes giving a deceleration of 6 m/s^2 to his scooter . How far will it travels before stopping?

Answer 1

Answer:

Given:

Initial velocity of scooter = 43.2 km/hr

Deceleration = 6 m/s²

To find:

Distance travelled before stopping

Concept:

Whenever brakes are applied in a moving vehicle( having an initial velocity) , the object experiences deceleration . The object travels some distance before coming to rest.

It can be calculated by using Equations of Kinematics since the deceleration is constant.

Conversion:

43.2 km/hr = 12 m/s

Calculation:

${v}^{2} = {u}^{2} + 2as$

$= > {0}^{2} = {(12)}^{2} + 2( - 6)s$

$= > 12s = 144$

$= > s = 12 \: metres$

So final answer :

$\boxed{ \red{ \huge{ \sf{ \bold{ s = 12 \: metres}}}}}$

Answer 2

Question :-

A person traveling at 43•2 km/h applies the brakes giving a deceleration of 6 m/s^2 to his scooter . How far will it travels before stopping?

Answer :-

Person will cover 12 metre before stopping .

To Find :-

→ Distance covered by person before stopping .

Solution :-

$\pink{ \boxed{\bf{{ \red{G}iv \green{en}}\begin{cases}\sf{ \green{initial \: velocity(u) \red{ \: = 43.2 }}\: \: \frac{km}{hr} }\\ \sf{ \blue{deceleration \: (a) =6} \: \frac{ m}{ {s}^{2} } }\\\sf{ \green{covered \: distance} \: (s) =? \: } \end{cases}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }\: }$

Here we will have to apply 3rd Equation of motion , i.e,

$\implies \boxed{ \sf{ \red{{v}^{2}} \ = {u}^{2} + \green{ 2as }\: }}$

Here final velocity (v) become zero because of stopping of motion .

Firstly convert km/hr into m/s

For this conversion multiply by 5/18 for m/s

$\to \sf{ u \: = \red{43.2 \times \frac{5}{18} \:} \frac{m}{s} } \\ \\ \to \sf{ u \: \green{ = 12 \: \frac{m}{s} }}$

(Approximately initial velocity )

Putting these given values in equation of motion

Note :-

→ Deceleration or retardation is negative

$\implies \sf{ 0 = \green{ {(12)}^{2} }\red{+ 2 \times( - 6) }\times s \: } \\ \\ \implies \: \sf{ 144 = 12 \: \times s \: } \\ \\ \implies \green{\boxed{ \sf{ s \: \red{ = 12 \: m} \: }}}$

That person will stop after 12 metre .