Question

A sedan car of mass 200 kg is moving in a certain velocity. It's brought to rest by application of breaks within 20m when average resistance offered is 500n.

Answer 1

stopping distance =
$m {u}^{2} \div 2force$
( 200 × u × u )/ 2 × 500 = 20

200×u×u /1000 = 20

200×u×u =20×1000

u×u = 20000/200

u×u =100

u = 10m/s

for applied retardation
${v}^{2} - {u}^{2} = 2as$
${0}^{2} - {10}^{2} = 2a \times 20$
100 = 2a ×20
5 = 2a
a = 5/2
$a = 2.5m {s}^{ - 2}$

Answer 2

According to Newton's second law of motion,

F=ma

a=F/m

a=500/200

a=5/2m/s

We know that,

V^2 - U^2 = 2as

0^2 - u^2= 2(5/2)(20)

u^2= 100

u = √100

u = 10