JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 14)
A particle is moving in a straight line. The variation of position '$$x$$' as a function of time '$$t$$' is given as $$x=\left(t^3-6 t^2+20 t+15\right) m$$. The velocity of the body when its acceleration becomes zero is :
6 m/s
10 m/s
8 m/s
4 m/s
Explanation
The position equation is given by:
$$x = t^3 - 6t^2 + 20t + 15$$
First, compute the velocity $$v$$ by differentiating the position function with respect to time:
$$ \frac{d x}{d t} = v = 3t^2 - 12t + 20 $$
Next, compute the acceleration $$a$$ by differentiating the velocity function with respect to time:
$$ \frac{d v}{d t} = a = 6t - 12 $$
We need to find the time $$t$$ when the acceleration is zero:
$$ 6t - 12 = 0 $$
$$ t = 2 \, \mathrm{sec} $$
Now, find the velocity at $$t = 2 \, \mathrm{sec}$$:
$$ v = 3(2)^2 - 12(2) + 20 $$
$$ v = 8 \, \mathrm{m/s} $$
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