Question

A car moves at a speed of 40km/h. It is stopped by applying brake which produces a uniform acceleration of -0.5m/s square.how much distance will the vehicle move before coming to stop ?

Answer 1

Answer:

Given :-

• A car moves at a speed of 40 km/h.
• It is stopped by applying brake which produces a uniform acceleration of - 0.5 m/s².

To Find :-

• What is the distance will the vehicle move before coming to stop.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

Solution :-

First, we have to convert initial velocity km/h to m/s :

$\implies \sf Initial\: Velocity =\: 40\: km/h$

$\implies \sf Initial\: Velocity =\: 40 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Initial\: Velocity =\: \dfrac{200}{18}\: m/s$

$\implies \sf\bold{\purple{Initial\: Velocity =\: 11.11\: m/s}}$

Now, we have to find the distance will the vehicle move before coming to stop :

Given :

• Initial Velocity (u) = 11.11 m/s
• Final Velocity (v) = 0 m/s
• Acceleration (a) = - 0.5 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf (0)^2 =\: (11.11)^2 + 2(- 0.5)s$

$\longrightarrow \sf 0 \times 0 =\: 11.11 \times 11.11 + (- 1)s$

$\longrightarrow \sf 0 =\: 123.43 - 1s$

$\longrightarrow \sf 0 - 123.43 =\: - 1s$

$\longrightarrow \sf {\cancel{-}} 123.43 =\: {\cancel{-}} 1s$

$\longrightarrow \sf 123.43 =\: 1s$

$\longrightarrow \sf \dfrac{123.43}{1} =\: s$

$\longrightarrow \sf\bold{\red{s =\: 123.43\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: will\: the\: vehicle\: move\: before\: coming\: to\: stop\: is\: 123.43\: m\: .}}}}$