Question

The brakes applied to a train moving at 34km/h produces a
retardation of 6m/s 2 . What distance will it cover before coming to a
stop?

Answer 1

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 34 km/h
• Acceleration (a) = –6 m/s² (as retardation is +6 m/s²)
• Final velocity (v) = 0 (as it comes to rest)

We are asked to calculate the distance covered by the body before coming to rest.

Let's first convert initial velocity in m/s.

$\longmapsto \rm { u = 34 \; kmh^{-1} }$

$\longmapsto \rm { u = \Bigg ( 34 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto \rm { u = \Bigg ( \dfrac{170}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto \bf { u = 9.44 \; ms^{-1} }$

$\rule{200}2$

Now, by using the third equation of motion :

$\longmapsto \rm {v^2 - u^2 = 2as }$

$\longmapsto \rm {(0)^2 - (9.44)^2 = 2 \times (-6) \times s }$

$\longmapsto \rm {0- 89.11 = -12 s }$

$\longmapsto \rm {- 89.11 = -12 s }$

$\longmapsto \rm {\dfrac{- 89.11}{-12} = s }$

$\longmapsto \bf { 7.42 \; m= s }$

Therefore, it will cover 7.42 m before coming to rest.

More Information :

Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time