Question

Plz answer this it’s urgent

Answer 1

given:-

$u = 72 \: kmh^{ - 1} = 72 \times \frac{5}{18} = 20 \: ms ^{ - 1}$

$s \: = 100 \: m$

ACT third equation of motion,

${v}^{2} = {u}^{2} + 2as$

$0 = {20}^{2} + 2 \times a \times 100$

$2 \times a \times 100 = - 400$

$a = - 2 \: ms ^{ - 2}$

and,

$v = u + at$

$0 = 20 + ( - 2) \times t$

$- 20 = - 2 \times t$

$t = 10 \: sec$

so,

the retardation of car is -2 m/s^2

and the time to stop is 10 sec

Answer 2

Answer:

72 kmph=5/18×72

              =20 m/s

∴v²=u²+2as

 0²=20²+2a×100

 200a=-400

        a=-2 m/s²

v=u+at

0=20-2t

2t=20

∴t=10 s

  $Therefore, Retardation= -2 m/s^2\\ Time to stop the car=10 s$