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Question

Physics Question on Motion in a straight line

A car accelerates from rest at constant rate for first 10s10\, s and covers a distance xx. It covers a distance yy in next 10s10\,s at the same acceleration. Which of the following is true?

A

x = 3y

B

y = 3x

C

x = y

D

y = 2x

Answer

y = 3x

Explanation

Solution

Since car accelerates from rest initial velocity is zero. From equation of motion, we have s=ut+12at2s=u t+\frac{1}{2} a t^{2} where uu is initial velocity, tt is time and a is acceleration. Since car accelerates from rest u=0,t=10su=0,\,\, t=10\, s s=0+12×a×(10)2=50a\therefore s=0+\frac{1}{2} \times a \times(10)^{2}=50 a..(i) Also, v=u+atv=u +a t where, vv is final velocity. \therefore Velocity after 10s10\, s is v=0+a×10v=0+a \times 10 v=10a=10×s50v=10\, a=10 \times \frac{s}{50}..(ii) In the next 10s10\,s car moves with constant acceleration and with initial velocity vv. s=vt+12at2\therefore s'=v t+\frac{1}{2} a t^{2} =s50×10×10+12×s50×100=3s=\frac{s}{50} \times 10 \times 10+\frac{1}{2} \times \frac{s}{50} \times 100=3\, s Given, s=xs=x and s=ys'=y y=3x\therefore y=3\, x