Question:
A car is standing $200\, m$ behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2 \, m/s^2$ and the car has acceleration $4 \, m/s^2$. The car will catch up with the bus after a time of :
A car is standing $200\, m$ behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2 \, m/s^2$ and the car has acceleration $4 \, m/s^2$. The car will catch up with the bus after a time of :
Updated On: Oct 1, 2024
- $\sqrt{110} s $
- $\sqrt{120} s $
- $10 \sqrt{2} s$
- $15\, s$
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The Correct Option is C
Solution and Explanation
$a_{CB} = 2 \,m/sec ^2$
$200=\frac{1}{2}\times2t^{2}$
$t=10\sqrt{2}$ second
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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.