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Question

Physics Question on Motion in a straight line

A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance s1s_1 in the first 10 seconds and distance s2s_2 in the next 10 seconds, then

A

s2s_2 = s1s_1

B

s2s_2 = 2s12 s_1

C

s2s_2 = 3s13 s_1

D

s2s_2 = 4s14 s_1

Answer

s2s_2 = 3s13 s_1

Explanation

Solution

Let a be the constant acceleration of the particle. Then s=ut+12at2s = ut + \frac{1}{2} at^2 or s1=0+12×a×(10)2=50as_1 = 0 + \frac{1}{2} \times a \times (10)^2 = 50 a and s2=[0+12a(20)2]50a=150as_2 = \left[ 0 + \frac{1}{2} a (20)^2\right] - 50a = 150 a s2=3s1\therefore \, s_2 = 3s_1