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Chapter 2 Motion In One Dimension | Exercise 2A

Question 46

A car is moving in a straight line with speed 18km/h. It is stopped in 5s by applying the brakes. Find: (i) the speed of the car in m/s, (ii) the retardation and (iii) the speed of the car after 2s of applying the brakes.

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Given: t=5s, intial velocity=18km/hr

(i) Expressing 18km/hr to m/s

18 \mathrm{km} \mathrm{h}^{-1}=\frac{18 \times 1000 \mathrm{m}}{60 \times 60}=5 \mathrm{ms}^{-1}

(ii) As the car comes to a halt, the final velocity is 0

Retardation is negative acceleration,

\text { Retardation }=\frac{v-u}{t}=\frac{0-5}{5}=-1 \mathrm{m} / \mathrm{s}^{2}

Retardation is 1m/s^2

(iii) If ‘v’ is the speed of the car after 2s of applying brakes, then acceleration is

\begin{array}{c} \text { Acceleration }=\frac{v-\mathrm{u}}{\mathrm{t}} \\ -1=\frac{\mathrm{v}-5}{2} \\ \mathrm{v}-5=-2 \\ \mathrm{v}=3 \mathrm{m} / \mathrm{s} \end{array}

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Question 46

A car is moving in a straight line with speed 18km/h. It is stopped in 5s by applying the brakes. Find: (i) the speed of the car in m/s, (ii) the retardation and (iii) the speed of the car after 2s of applying the brakes.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given: t=5s, intial velocity=18km/hr

(i) Expressing 18km/hr to m/s

18 \mathrm{km} \mathrm{h}^{-1}=\frac{18 \times 1000 \mathrm{m}}{60 \times 60}=5 \mathrm{ms}^{-1}

(ii) As the car comes to a halt, the final velocity is 0

Retardation is negative acceleration,

\text { Retardation }=\frac{v-u}{t}=\frac{0-5}{5}=-1 \mathrm{m} / \mathrm{s}^{2}

Retardation is 1m/s^2

(iii) If ‘v’ is the speed of the car after 2s of applying brakes, then acceleration is

\begin{array}{c} \text { Acceleration }=\frac{v-\mathrm{u}}{\mathrm{t}} \\ -1=\frac{\mathrm{v}-5}{2} \\ \mathrm{v}-5=-2 \\ \mathrm{v}=3 \mathrm{m} / \mathrm{s} \end{array}

Our top 5% students will be awarded a special scholarship to Lido.

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