Motion in a Plane - Result Question 23
25. The $x$ and $y$ coordinates of the particle at any time are $x=5 t-2 t^{2}$ and $y=10 t$ respectively, where $x$ and $\mathbf{y}$ are in meters and $t$ in seconds. The acceleration of the particle at $t=2 s$ is
(a) $5 m / s^{2}$
(b) $-4 m / s^{2}$
(c) $-8 m / s^{2}$
(d) 0
[2017]
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Answer:
Correct Answer: 25. (b)
Solution:
- (b) Given:
$x=5 t-2 t^{2} \quad y=10 t$
$v_x=\frac{d x}{d t}=5-4 t \quad v_y=\frac{d y}{d t}=10$
$a_x=\frac{d v_x}{d t}=-4 \quad a_y=\frac{d v_y}{d t}=0$
$\vec{a}=a_x i+a_y j \quad \vec{a}=-4 i m / s^{2}$
Hence, acceleration of particle at $(t=2 s)=-4 m / s^{2}$
