Question

A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?​

Answer 1

Explanation:

Initial velocity of the car, u = 126 km/h = 35 m/s

Final velocity of the car, v = 0

Distance covered by the car before coming to rest, s = 200 m

Retardation produced in the car = a

From third equation of motion, a can be calculated as:

v2 - u2 = 2as

(0)2 - (35)2 = 2 × a × 200

a = - 35 × 35 / 2 × 200 = - 3.06 ms-2

From first equation of motion, time (t) taken by the car to stop can be obtained as:

v = u + at

t = (v - u) / a = (- 35) / (-3.06) = 11.44 s

Answer 2

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

${\bold{\therefore Retardation=3.06\:m/s^{2}}}$

${\bold{\therefore Time=11.4\:sec}}$

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

• In the given question information given about a car which is travelling at a constant speed and after sometime it came to stop within a distance of 200m.

• We have to find the retardation of car and time taken.

$\underline \bold{Given : } \\ \implies Initial \: speed(u) = 126 \: kmh \\ \\ \implies Final \: speed(v) = 0 \\ \\ \implies Distance(s) = 200 \: m \\ \\ \underline \bold{To \: Find : } \\ \implies retardation = ? \\ \\ \implies Time(t) = ?$

• According to given question :

$\bold{Initial \: speed = 126 \: kmh = 35 \: ms } \\ \\ \bold{By \: second \:equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {0}^{2} = {35}^{2} + 2 \times a \times 200 \\ \\ \implies - 1225 = 400a \\ \\ \implies a = \frac{ \cancel {- 1225}}{ \cancel400} \\ \\ \bold{\implies a = - 3.06 \: m/ {s}^{2} } \\ \\ \bold{\therefore retardation \: of \: 3.06 \: m /{s}^{2} } \\ \\ \bold{by \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 0 = 35 + (- 3.06 )\times t \\ \\ \implies t = \frac{ \cancel{- 35}}{ \cancel{- 3.06}} \\ \\ \bold{\implies t = 11.4 \: sec}$