Question:
If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three second, the time of the travel is
If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three second, the time of the travel is
Updated On: Jul 28, 2022
- 6 s
- 5 s
- 4 s
- 3 s
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The Correct Option is B
Solution and Explanation
The distance covered in $3 \,s$,
$s_3$ $=ut$ $+\frac{1}{2}\,gt^2$
or $s_3=0 \times t +\frac{1}{2}9.8 \times$ $3^2$
or $s_3 =44.1 m$
The distance covered in last second $s_1=u+\frac{1}{2}g(2t-1)$
$44.1=0+\frac{1}{2}\times 9.8 (2t-1)$ =4.9(2t-1)
As both the distances are equal then
$4.9 (2t -1)=44.1$
$2t-1=\frac{44.1}{4.9}=9$
or $2t =9+ 1 = 10$ or $t=5 \,s$
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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:
- 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.
- 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.