Question

a car is moving at a uniform speed of 72 km/h.the driver sees a child at a distance of 50m.he applies a breaks to stop the car just before the child.calculate the acceleration

Answer 1

The answer of this question is 51.84m/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 ;

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

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