A car moves with a uniform velocity of 40 $ms^{-1}$ for 5 s. It comes to rest in the next 10 s with uniform deceleration. | ClassNotes

Jun 19, 2022

A car moves with a uniform velocity of 40 $ms^{-1}$ for 5 s. It comes to rest in the next 10 s with uniform deceleration.

Find:

Difficulty: Easy

Solution:

Initial velocity = Vi = 40 $ms^{-1}$

Time = t = 5 s

Final velocity = Vf = 0 $ms^{-1}$

Time = 10 s

(I) deceleration a =?

(II)total distance S =?

Vf = Vi + at

Or at = Vf - Vi

a = $\frac{Vf - Vi}{T}$

a = $\frac{0 – 40}{10} $

a = $-4 ms^{-2}$

Total distance travelled = S = S1 + S2

By using this relation

S1 = Vt

S1 = $40 \times 5$

S1 = 200 m ……………………. (i)

Now by using 3rd equation of motion

2aS_${2} = Vf^{2} – Vi^{2}$

S2 = $Vf^{2} –\frac{Vi2}{2a}$

S2 = $(0)2 – (40)2/2\times (-4)$

S2 =$\frac{-1600}{-8}$

S2 = 200 m …………………………… (ii)

From (i) and (ii) we get;

S = S1 + S2

Or S = 200 m + 200 m

S = 400 m