Question

A car moves at a speed of 40km/h, it is stopped by applying brakes which produces a uniform acceleration of -0.6m/s2. how much distance will the vehicle moves before coming to stop ???????​

Answer 1

Given :

• Initial Velocity of a car, u = 40km/hr=11.1m/s
• Final velocity ( car stopped ), v = 0
• Acceleration , a = -0.6m/s²

To Find :

The distance will the vehicle moves before coming to stop.

Formula used :

⇒Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Solution:

By using equation of motion

$\bf\:v{}^{2}=u{}^{2}+2aS$

$\sf\:0=(11.1)^2+2(-0.6)S$

$\sf\:(11.1)^2=2(0.6)S$

$\sf\:\dfrac{11.1\times11.1}{2\times0.6}=S$.

$\sf\dfrac{123.21}{1.2}=S$

$\sf\:S=102.67m$

Therefore ,The distance will the vehicle moves before coming to stop is 102.67m.

Answer 2

Answer:-

Given:-

• Initial velocity (u) = 40 km/h = 100/9 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = - 0.6 m/s²

To Find: Distance travelled before becoming stationary.

We know,

2as = v² - u²

where,

• a = Acceleration,
• s = Distance travelled,
• v, u = Final and initial velocity respectively.

So,

s = (v² - u²)/2a

→ s = - u²/2a

→ s = - (100/9 m/s)²/2(- 0.6 m/s²)

→ s = (10000/81 m²/s²)/(1.2 m/s²)

→ s = (10000/81 m)/(12/10)

→ s = (10000/81 m)/(5/6)

→ s = 102.88 m

∴ Before becoming stationary, it covered approximately 103 metres.