Question

A motor car slows down from 72 km/h to 36 km/h over at a distance of
25 m.If the brakes applied with the same force, find time taken by it to
come to rest.​

Answer 1

Answer:

Time taken to stop car = 1.7 s

Step by step explanations :

Given that,

A motor car slows down from 72 km/h to 36 km/h over at a distance of

25 m

here,

Initial velocity of the car = 72 km/h

= 72 × 1000 /3600 m/s

= 20 m/s

final velocity of the car = 36 km/h

= 10 m/s

distance travelled between = 25 m

let the acceleration of the car be a

now we have,

Initial velocity(u) = 20 m/s

final velocity(v) = 10 m/s

distance travelled(s) = 25 m

acceleration = a

so,

by the equation of motion,

v² = u² + 2as

putting the values,

10² = 20² + 2a(25)

50a + 400 = 100

50a = 100 - 400

50a = -300

a = -300/50

a = -6 m/s²

now,

after that brakes are applied with the same force

here,

initial velocity(u) = 10 m/s

final velocity(v) = 0 m/s [It will stop]

acceleration(a) = - 6 m/s²

let the time taken to stop be t

by the equation of motion,

v = u + at

putting the values,

0 = 10 + (-6)t

-6t = -10

t = 10/6

t = 1.7 s

so,

Time taken to stop car = 1.7 s

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=3.34\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a motor car slows down from 72 km/h to 36 km/h over at a distance of

25 m.If the brakes applied with the same force.

• We have to find time taken by it to

come to rest.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 72 \: km/h \\ \\ \implies Final \: velocity(v) = 36 \: km/h \\ \\ \implies Distance(s) = 25 \: m \\ \\ \underline\bold{To\:Find:}\\\implies time \: taken(t) = ?$

• According to given question :

$\bold{u = 72 \times \frac{5}{18} =20 \: m/s }\\ \\ \bold{v = 36 \times \frac{5}{18} = 10 \: m/s} \\ \\ \bold{by \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {10}^{2} = {20}^{2} + 2 \times a \times 25 \\ \\ \implies 100 - 400 = 50 \times a \\ \\ \implies - 300 = 50 \times a \\ \\ \implies a = \frac{ - 300}{50} \\ \\ \bold{\implies a = - 6\: m /{s}^{2} } \\ \\ \bold{for \: get \: car \: in \: rest}\\ \implies v = 0 \: m/s \\ \\ \bold{by \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 0 = 20 + ( - 6) \times t \\ \\ \implies - 20 = - 6 \times t \\ \\ \implies \cancel - 20 = \cancel- 6 \times t \\ \\ \implies t = \frac{20}{6} \\ \\ \bold{ \implies t = 3.33</p><p>4\: sec} \\ \\ \bold{\therefore time \: taken \: to \: take\:car\:in\:rest \:is \: 3.34\: sec}$