Motion in a Straight Line - Result Question 24
25. The displacement of a particle is represented by the following equation : $s=3 t^{3}+7 t^{2}+5 t+8$ where $s$ is in metre and $t$ in second. The acceleration of the particle at $t=1 s$ is
(a) $14 m / s^{2}$
(c) $32 m / s^{2}$
(b) $18 m / s^{2}$
(d) zero
[2000]
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Answer:
Correct Answer: 25. (c)
Solution:
- (c) Displacement
$s=3 t^{3}+7 t^{2}+5 t+8$
Velocity $=\frac{d s}{d t}=9 t^{2}+14 t+5$
Acceleration $=\frac{d^{2} s}{d t^{2}}=18 t+14 \quad \square=v _{\max }(\frac{1}{\alpha}+\frac{1}{\beta})=v _{\max }(\frac{\alpha+\beta}{\alpha \beta})$ Acceleration at $(t=1 s)$ $=18 \times 1+14=18+14=32 m / s^{2}$
