Question:

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 ms^{-1}$ to $20ms^{-1}$ while passing through a distance 135 m in t second. The value of t is

Updated On: Apr 7, 2024
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The Correct Option is D

Solution and Explanation

Let $u$ and $v$ be the first and final velocities of particle and $a$ and $s$ be the constant acceleration and distance covered by it. From third equation of motion
$v^2=u^2+2as$
$\Rightarrow$ $(20)^2=(10)^2+2a \times 135$
or $a=\frac{300}{2 \times 135}=\frac{10}{9}ms^{-2}$
Now using first equation of motion,
v=u+at
or $t=\frac{v-u}{a}=\frac{20-10}{(10/9)}=\frac{10 \times 9}{10}=9s$
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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.