Question

a car running at 72 km per hour is slow down to 18 km per hour by the application of break over a distance of 20 metres then find Total distance covered before it comes to rest​

Answer 1

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

$\green{\tt{\therefore{Total\:distance\=41.34\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial\: velocity(u) = 72 \: km/h \\ \\ \tt : \implies Final\:velocity(v)=18 \: km/h\\ \\ \tt: \implies Distance(s) = 20\:m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Total\:distance\:covered\:before\:it\:comes\:to\:rest = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 72 \times \frac{5}{18} = 20 \: m/s \\ \\ \tt \circ \:Final\:velocity = 18\times\frac{5}{18}=5 \: m/s \\ \\ \tt \circ \:Distance = 20 \: m$

$\bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {5}^{2} = {20}^{2} + 2 \times a \times 20 \\ \\ \tt: \implies 25 = 400 + 40 \times a \\ \\ \tt: \implies 25 - 400 = 40 \times a \\ \\ \tt: \implies a = \frac{ - 375}{40} \\ \\ \tt: \implies a = - 9.375 \: {ms}^{2}$

$\bold{For \: car \: in \: rest} \\ \tt \circ \: Final \: velocity = 0 \: ms \\ \\ \tt \circ \: Initial \: velocity = 20 \: ms \\ \\ \tt \circ \: Acceleration = - 9.375 \: m {s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {20}^{2} + 2 \times (- 9.375) \times s \\ \\ \tt: \implies - 400 = - 18.75 \times s \\ \\ \tt: \implies \frac{400}{18.75} = s \\ \\ \green{\tt: \implies s = 21.34 \: m} \\ \\ \bold{For \: total \: distance} \\ \tt: \implies Total \: distance = 20 + 21.34 \\ \\ \green{\tt: \implies Total \: distance =41.34 \: m}$