Question:

A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the $4^{th}$ second compared to that in the $3^{rd}$ second is

Updated On: Apr 19, 2024
  • $33\%$
  • $40\%$
  • $66\%$
  • $77\%$
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The Correct Option is B

Solution and Explanation

We know that
$S_{n t h}=u+\frac{1}{2} a(2 n-1) $
$S_{3 \text { rd }}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a ($ for $ n=3 s)$
$S_{4 t h}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a ($ for $n=4 s )$
So, the percentage increase
$=\frac{S_{4 h }-S_{3 rd }}{S_{3 rd }} \times 100$
$=\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100$
$=\frac{\frac{2 a}{2}}{\frac{5}{2} a} \times 100$
$=2 \times 20=40 \%$
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Concepts Used:

Motion in a straight line

The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion. 

Types of Linear Motion:

Linear motion is also known as the Rectilinear Motion which are of two types:

  1. Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
  2. Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.