Question

A car is moving with a velocity of 72 km/h. Its velocity is reduced to 36 km/h after covering a distance of 200 m. Calculate retardation in cm/s^2. If the same retardation is
continued, what further distance does the car move before coming to rest?
Ans: 75 cm/s^2 ,66.67 m
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Answer 1

Aɴsᴡᴇʀ

★ case I :

• initial velocity, u = 72km/h
• final velocity, v = 36km/h
• distance, s = 200m

first, we have to convert the unit of velocity from km/h to m/s

• for converting km/h to m/s multiply the velocity by 5/18

$\bold{\implies u = \cancel{72 }\times \frac{5}{ \cancel{18}} m {s}^{ - 1}} \: \: \: \: \\ \\ \bold{ \implies u = 4 \times 5 \: m {s}^{ - 1} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies u = 20 \: m {s}^{ - 1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

and

$\bold{ \implies \: v = \cancel{ 36 }\times \frac{5}{ \cancel{18}} \: m {s}^{ - 1} } \: \\ \\ \bold{ \implies v = 2 \times 5 \: m {s}^{ - 1} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies v = 10 \: m {s}^{ - 1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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so now we have,

• initial velocity, u = 20 m/s
• final velocity, v = 10 m/s
• distance, s = 200m
• retardation = ?

now, by third equation of motion:

$\boxed{ \red{\bold{{v}^{2} = {u}^{2} + 2as}}}$

$\bold{ : \implies {(10)}^{2} = {(20)}^{2} + 2 \times a \times 200} \\ \\ \bold{ : \implies100 = 400 + 400a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies 400a = 100 - 400 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a = \frac{ - 3 \cancel{00}}{4 \cancel{00}} } \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a = - \frac{3}{4} \: m {s}^{ - 2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now, convert the unit of acceleration,a into cm/s²

$\bold{ : \implies a = \frac{ - 3 \times \cancel{100}}{ \cancel{4}}cm \: {s}^{ - 2} } \\ \\ \bold{: \implies a = - 3 \times 25 \: cm \: {s}^{ - 2} } \: \: \\ \\ \bold{ : \implies \boxed{ \bold{ a = - 75 \: cm \: {s}^{ - 2} }} } \: \: \: \: \: \: \: \:$

[negative sign is indication of retardation.

as we know that, negative acceleration is called retardation]

so, retardation produced by the car is 75cm/s²

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★ case II :

• initial velocity, u = 10m/s
• final velocity, v = 0m/s
• acceleration, a = -75cm/s² = -3/4m/s²
• distance, s = ?

by third equation of motion:

$\boxed{ \red{\bold{{v}^{2} = {u}^{2} + 2as}}}$

$\bold{: \implies \: {(0)}^{2} = {(10)}^{2} + \cancel{ 2} \times \frac{ - 3}{ \cancel{4}} \times s } \\ \\ \bold{ : \implies0 = 100 + \frac{ - 3}{2} \times s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \bold{ : \implies - \frac{3s}{2} = - 100 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: s = \frac{ - 100 \times - 2}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies s = \cancel{\frac{ 200}{3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies s = 66. 6\bar{6}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \boxed{ { \bold{s =66. 67 \: m}}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, distance covered by the car before coming to rest is 66.67 m

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☯︎ additional information :-

• first equation of motion : at = v - u
• second equation of motion : s = ut + ½ at²
• third equation of motion : v² = u² + 2as