Question

a car is moving at a uniform speed of 72km/hr. the driver sees a child at a distance of 50km. he applies brakes to stop the car just in front of the child. calculate the acceleration of the car.

Answer 1

deceleration = u^2/2s = 72*72/(2*50) = 51.84 km/h^2
actually your question is wrong I guess because you cannot see anything that is 50 km far and the maximum distance we can actually see before the earth and sky meets visually is 5 km
so it must be 50 m
72kmph = 20 m/s
so a = 20*20/(2*50) = 4 m/s^2

Answer 2

$\bf{\underline{\underline{Answer:}}}$

• $\tt The \:acceleration = - 4 \:m\: s^{-2}$

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

• $\tt Car\: uniform\: speed = 72 \:km\: h ^{-1}$

• $\tt Driver \:sees \:a \:child\: at \:distance\: = \:50 \:m$

$\bold{\underline {To \:find :}}$

• $\tt The \:acceleration = ?$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

$\tt Initial\: velocity,\: u \:= \:72 \:km\: h^{-1}\:\:\:\:\:[\sf Given]\\ \\= \tt \dfrac{72 \times 1000}{60 \times 60}\\ \\ = \tt \dfrac{720}{36}\\ \\ \tt = 2 \:m\:s^{-1}$

$\rule{200}{2}$

• Distance, s = 50 m

• Final velocity, v = 0

• Acceleration = a =?

$\bigstar \boxed{\tt v^2 = u^2 + 2as}$

Substituting the values in the above equation, we get ;

$:\implies \tt 0 = 20^2 + 2 \times a \times 50 \\ \\ : \implies \tt 100a = - 400 \\ \\ :\implies \tt a = \dfrac{-400}{100}\\ \\ :\implies \tt{\boxed {\green {a =\tt - 4\: m\: s^{-2}}}}$

${\bf {Thus, \:The\: acceleration}} = \tt{\boxed {\green {a =\tt - 4\: m\: s^{-2}}}}$