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2.Motion in Straight Line
hard
A particle is moving in one dimension (along $\mathrm{x}$ axis) under the action of a variable force. It's initial position was $16 \mathrm{~m}$ right of origin. The variation of its position ( $\mathrm{x}$ ) with time ( $\mathrm{t})$ is given as $\mathrm{x}=-3 \mathrm{t}^5+18 \mathrm{t}^2+16 \mathrm{t}$, where $\mathrm{x}$ is in $\mathrm{m}$ and $\mathrm{t}$ is in $\mathrm{s}$. The velocity of the particle when its acceleration becomes zero is________ $\mathrm{m} / \mathrm{s}.$
A$50$
B$52$
C$57$
D$60$
(JEE MAIN-2024)
Solution
$x=3 t^3+18 t^2+16 t$
$v=-9 \mathrm{t}^2+36+16$
$a=-18 \mathrm{t}+36$
$a=0 \text { at } \mathrm{t}=2 \mathrm{~s}$
$v=-9(2)^2+36 \times 2+16$
$v=52 \mathrm{~m} / \mathrm{s}$
$v=-9 \mathrm{t}^2+36+16$
$a=-18 \mathrm{t}+36$
$a=0 \text { at } \mathrm{t}=2 \mathrm{~s}$
$v=-9(2)^2+36 \times 2+16$
$v=52 \mathrm{~m} / \mathrm{s}$
Standard 11
Physics
Similar Questions
Match the following columns.
| Colum $I$ | Colum $II$ |
| $(A)$ $\frac{dv}{dt}$ | $(p)$ Acceleration |
| $(B)$ $\frac{d|v|}{dt}$ | $(q)$ Magnitude of acceleration |
| $(C)$ $\frac{dr}{dt}$ | $(r)$ Velocity |
| $(D)$ $\left|\frac{d r }{d t}\right|$ | $(s)$ Magnitude of velocity |
easy
