Question

a moving car along a straight line highway with speed of 126/hrs brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?​

Answer 1

$\: \: \huge\mathbb\colorbox{black}{\color{white}{AnSwEr}}$

$\sf \pink{Given} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \green{Velocity \: of \: car \: v = 126 km/h = 35 m/s}$

$\sf \green{Displacements \: 200 \: m} \: \: \: \: \\ \sf \pink{Final \: velocity \: v = 0 m/s}$

$\sf {Apply \: second \: kinematic \: equation \: to} \\ \sf calculate \: retardation \: a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: \:(v²-u² = 2as) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: a = \frac{ {v}^{2} - {u}^{2} }{2s} = \frac{0 - {35}^{2} }{2 \times 200} = - 3.0625 \: {m/s}^{2}$

$\sf \green{Apply \: first \: kinematic \:} \\ \sf \green {equation \: to \: calculate \: timet}$

$\sf \: v = u + at$

$\sf \: = > \: t = \frac{v - u}{a} = \frac{0 - 35}{3.0625} = 11.43sec$

$\sf \: Hence, retardation \: is \: 3.06 m/s² \: and \: time \\ \sf \: take \: to \: stop \: is \: 11.4 sec \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \green \: ans = > \: 11.4sec$

hope it was helpful to you