Question

a car travelling at a speed of 15m/s in brought.is rest in 90s.find the retardation and distance travelled by the ear before coming in rest
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Answer 1

Given :-

• Initial velocity, u = 15 m/2
• Final velocity, v = 0 m/s
• Time, t = 90s

To Find :-

• Retardation (a).
• Distance travelled by the car before coming to rest (s).

Solution :-

First equation of motion :-

$\bf\boxed{\bf v=u+at}$

$:\implies \sf 0 = 15+ 90 a$

$:\implies \sf 90a=-15$

$:\implies \sf a = \dfrac{-15}{90}$

$:\implies \sf a = \dfrac{-1}{6} \ m/s^2$

Retardation = -1/6 m/s²

Third equation of motion :-

$\boxed{\bf v^2= u^2+2as}$

$:\implies \sf 0^2=15^2+ 2 \times \dfrac{-1}{6} \times s$

$:\implies \sf 0 = 225-\dfrac{s}{3}$

$:\implies \sf \dfrac{s}{3} = 225$

$:\implies \sf s = 675 m$

Distance travelled by car before coming to rest = 675m

Answer 2

Answer:

Retardation= 0.106m/s^2

Distance travelled by the car before coming to rest=675m

Explanation:

u= 15m/s

v= 0m/s

t=90s

r=v-u/t

r=0.106m/s^2

Distance S= 1/2x(u+v)xt

S=15/2x90

S=675m