Question:

A block of mass $5 \,kg$ is moving horizontally at a speed of $ 1.5\,m{{s}^{-1}} $ . A vertically upward force $5 N$ acts on it for $4 s$. What will be the distance of the block from the point where the force starts acting?

Updated On: Jul 12, 2022
  • 2 m
  • 6 m
  • 8 m
  • 10 m
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The Correct Option is D

Solution and Explanation

Upward acceleration $=\frac{5}{5}=1 \,m / s ^{2}$ Upward distance covered in $4 s$ $y =\frac{1}{2} a t^{2}$ $=\frac{1}{2} \times 1 \times(4)^{2}=8 \,m$ Horizontal distance covered in $4 s$ $x =v t=1.5 \times 4=6\, m $ $\therefore s =\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+8^{2}} $ $=\sqrt{36+64}=10 \,m$
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Notes on Motion in a straight line

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.