
The velocity of an object moving rectilinearly is given as a function of time by $v=4t-3{{t}^{2}}$, where v is in m/s and t is in seconds. The average velocity of particle between t = 0 to t = 2 seconds is
A. 0
B. -2 m/s
C. – 4 m/s
D. +2 m/s
Answer
161.4k+ views
Hint: We are given that the velocity of an object moving rectilinearly, that is the particle is moving along a straight line. We have to find the average velocity of a particle. As the average velocity is the rate of change of distance with respect to time. So we integrate the given velocity and find out our correct answer.
Formula used:
$\text{Average velocity} = \dfrac{\text{distance covered}}{\text{time}}$
Complete step by step solution:
Given that as the velocity of an object moving rectilinearly.So this motion is represented by one co- ordinate only. The velocity of an object is given by,
$v = 4t-3{{t}^{2}}$…………………………………. (1)
We have to determine the average velocity of particle from t = 0 to t = 2 seconds
$\text{Average velocity} = \dfrac{\text{distance covered}}{\text{time}}$
$\Rightarrow {{v}_{avg}}=\dfrac{s}{({{t}_{2}}-{{t}_{1}})}$
$\Rightarrow v=\dfrac{ds}{dt}$
$\Rightarrow ds=vdt$
Where ds = distance covered in time dt
Therefore,
$s = \int_{0}^{2}{4t-3{{t}^{2}}}dt$
$\Rightarrow s=[2t^2-t^3]^2_0$
$\Rightarrow s=8-8=0$
Therefore, average velocity ${{v}_{avg}}=0$.
Hence, option A is correct.
Note: Students should understand that average velocity is not the same as the velocity. Velocity is defined as the displacement at a specific point of time. While the average velocity is the total change in the position or displacement for a given interim of time.
Formula used:
$\text{Average velocity} = \dfrac{\text{distance covered}}{\text{time}}$
Complete step by step solution:
Given that as the velocity of an object moving rectilinearly.So this motion is represented by one co- ordinate only. The velocity of an object is given by,
$v = 4t-3{{t}^{2}}$…………………………………. (1)
We have to determine the average velocity of particle from t = 0 to t = 2 seconds
$\text{Average velocity} = \dfrac{\text{distance covered}}{\text{time}}$
$\Rightarrow {{v}_{avg}}=\dfrac{s}{({{t}_{2}}-{{t}_{1}})}$
$\Rightarrow v=\dfrac{ds}{dt}$
$\Rightarrow ds=vdt$
Where ds = distance covered in time dt
Therefore,
$s = \int_{0}^{2}{4t-3{{t}^{2}}}dt$
$\Rightarrow s=[2t^2-t^3]^2_0$
$\Rightarrow s=8-8=0$
Therefore, average velocity ${{v}_{avg}}=0$.
Hence, option A is correct.
Note: Students should understand that average velocity is not the same as the velocity. Velocity is defined as the displacement at a specific point of time. While the average velocity is the total change in the position or displacement for a given interim of time.
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