A car moving with uniform velocity 60 km/hr retards in 10 sec and redu

1.	A car moving with uniform velocity 60 km/hr retards in 10 sec and reduces to 45 km/hr. find the retardation.
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{Retardation=\frac{2.5}{6}\:m/s^{2}\:or\:0.415\:m/s^{2}}}}\\$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies Initial \: velocity(u) = 60 \: km/h \\ \\ \tt: \implies Final \: velocity(u) =45 \: km/h \\\\ \tt: \implies Time(t) = 10 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Retardation(r) =?$ • ACCORDING to given QUESTION : $\tt \circ \: Intial \: velocity = 60 \times \frac{5}{18} = \frac{100}{6} \: m/s \\ \\ \tt \circ \: Final \: velocity = 45 \times \frac{5}{18} = \frac{75}{6} \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies \frac{75}{6} = \frac{100}{6} + a \times 10 \\ \\ \tt: \implies \frac{75}{6} - \frac{100}{6} = a \times 10 \\ \\ \tt: \implies 10 \times a = \frac{ - 25}{6} \\ \\ \tt: \implies a = \frac{ - 25}{6 \times 10} \\ \\ \green{\tt: \implies a = \frac{ - 2.5}{6} \: m/ {s}^{2} \: or \:-0.415\:m/s^{2}} \\ \\ \green{\tt \therefore Retardation \: of \: car \: is \: \frac{ 2.5}{6} \: m /{s}^{2} \:or\:0.415\:m/s^{2}}$

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

Answer»

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

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

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

$\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies Initial \: velocity(u) = 60 \: km/h \\ \\ \tt: \implies Final \: velocity(u) =45 \: km/h \\\\ \tt: \implies Time(t) = 10 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Retardation(r) =?$

• ACCORDING to given QUESTION :

$\tt \circ \: Intial \: velocity = 60 \times \frac{5}{18} = \frac{100}{6} \: m/s \\ \\ \tt \circ \: Final \: velocity = 45 \times \frac{5}{18} = \frac{75}{6} \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies \frac{75}{6} = \frac{100}{6} + a \times 10 \\ \\ \tt: \implies \frac{75}{6} - \frac{100}{6} = a \times 10 \\ \\ \tt: \implies 10 \times a = \frac{ - 25}{6} \\ \\ \tt: \implies a = \frac{ - 25}{6 \times 10} \\ \\ \green{\tt: \implies a = \frac{ - 2.5}{6} \: m/ {s}^{2} \: or \:-0.415\:m/s^{2}} \\ \\ \green{\tt \therefore Retardation \: of \: car \: is \: \frac{ 2.5}{6} \: m /{s}^{2} \:or\:0.415\:m/s^{2}}$