Question

Q.41 In which time interval does the retarding force act on the bady ? (a) t=0s to t= 10s (b) t= 8s to t= 12s (c) t= 10s to t= 12s (d) t = 12s to t= 14s​

Answer 1

Answer:

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Explanation:

f=−2v Velocity decreases to 37% if its initial value, if u=1, then v=32% of, 1 i.e v=0.32

m=10 kg

t=?

f=ma

−2v=10×a

a=dv/dt

dt

dv

=−0.2 v

−0.2 t=[logv]

1

0.37

−0.2 t=log(0.37)−log(1)

−0.2 t=−0.994

t=

0.2

0.994

t=

2

9.94

t=4.97

or t≈5 m/s