Question

The brakes applied to a car produce a negative acceleration of 10m/sec. If the car takes 5sec to stop after applying brakes. Calculate distance covered by a car before coming to rest

Answer 1

a= - 10 m/s²
t = 5 sec
v = 0 m/s
v= u +at
⇒ 0 = u + (-10)×5
⇒u = 50 m/s
now, s = ut + (1/2)at²
           = 50×5 + (1/2)×(-10)×(5)²
           = 250 -125 = 125 m

Answer 2

We have ,

$a = - 10m {s}^{2}$

$t = 5sec$

$v = 0ms$

From first eqn of motion ,

$v = u + at$

$0 = u + (- 10) \times 5$

$u = 50ms$

From second eqn of motion ,

$s = ut + \frac{1}{2} a {t}^{2}$

$50 \times 5 + \frac{1}{2} \times (-10) \times {5}^{2}$

$250 - 125$

$125m$