Question

A car moving with uniform velocity 60 km/hr retards in 10 sec and reduces to 45km/hr.Find the retardation.

Answer 1

ATQ,

initial velocity = u = 60km/hrs

final velocity = v = 45km/hrs

time = t = 10 secs

time is in seconds. so we have to first convert initial and final velocity in m/s

u = 60 × 5/18 = 300/18 m/s

v = 45 × 5/18 = 225/18 m/s

now, by using equation of motion we get,

➡ v = u + at

➡ v - u = at

➡ 225/18 - 300/18 = 10a

➡ -75/18 = 10a

➡ -75/18 × 1/10 = a

➡ a = -7.5/18 m/s^2

in km/hr = -7.5/18 × 18/5

a = -1.5km/hr^2

hence, the retardation is -15km/hr^2

Answer 2

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

$\green{\tt{\therefore{Retardation=\frac{2.5}{6}\:m/s^{2}\:or\:0.415\:m/s^{2}}}}$

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

$\green{\underline \bold{Given :}}$

$\tt: \implies Initial \: velocity(u) = 60 \: km/h$

$\tt: \implies Final \: velocity(u) =45 \: km/h$

$\tt: \implies Time(t) = 10 \: sec$

$\red{\underline \bold{To \: Find :}}$

$\tt: \implies Retardation(r) =?$

• Given Question

$\tt \circ \: Intial \: velocity = 60 \times \frac{5}{18} = \frac{100}{6} \: m/s$

$\tt \circ \: Final \: velocity = 45 \times \frac{5}{18} = \frac{75}{6} \: m/s$

$\tt: \implies v = u + at$

$\tt: \implies \frac{75}{6} = \frac{100}{6} + a \times 10$

$\tt: \implies \frac{75}{6} - \frac{100}{6} = a \times 10$

$\tt: \implies 10 \times a = \frac{ - 25}{6}$

$\tt: \implies a = \frac{ - 25}{6 \times 10}$

$\green{\tt: \implies a = \frac{ - 2.5}{6} \: m/ {s}^{2} \: or \:-0.415\:m/s^{2}}$

$\green{\tt \therefore Retardation \: of \: car \: is \: \frac{ 2.5}{6} \: m /{s}^{2} \:or\:0.415\:m/s^{2}}$