Question

A car moving with a speed of 54 km/hr is
bought to rest within 10 sec. by applying
breaks. Find the retardation of the cars and the
distance travelled by it.​

Answer 1

Answer:

retardation is 1.5 m/s2

speed =54km/hr

speed=54×5/18m/s

speed=15

a=v/t

a=15/10

a=1.5m/s2

Answer 2

Answer:

u = 54 kmh-1 = 15 ms-1

v = 0 ms-1

t = 10 s

Substituting the values in the eq.

v = u + at

0 = 15 + 10a

a = -1.5 ms-2

Substituting the values in the eq.

s = ut + 1/2at2

s = (15 x 10) + (1/2 x 100 x -1.5)

s = 75 m

Retardation = -1.5 ms-2

Distance travelled = 75 m