Kinematics Ques 46
- The position vector of particle changes with time according to the relation $\mathbf{r}(t)=15 t^{2} \hat{\mathbf{i}}+\left(4-20 t^{2}\right) \hat{\mathbf{j}}$. What is the magnitude of the acceleration (in $ms^{-2}$ ) at $t=1$ ?
(a) 50
(b) 100
(c) 25
(d) 40
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Answer:
Correct Answer: 46.(a)
Solution:
Formula:
- Position vector of particle is given as
$$ \mathbf{r}=15 t^{2} \hat{\mathbf{i}}+\left(4-20 t^{2}\right) \hat{\mathbf{j}} $$
Velocity of particle is
$$ \begin{aligned} \mathbf{v} & =\frac{d \mathbf{r}}{d t}=\frac{d}{d t} \quad\left[15 t^{2} \hat{\mathbf{i}}+\left(4-20 t^{2}\right) \hat{\mathbf{j}}\right] \\ & =30 t \hat{\mathbf{i}}-40 t \hat{\mathbf{j}} \end{aligned} $$
Acceleration of particle is
$$ \mathbf{a}=\frac{d}{d t}(\mathbf{v})=\frac{d}{d t}(30 t \hat{\mathbf{i}}-40 \hat{\mathbf{j}})=30 \hat{\mathbf{i}}-40 \hat{\mathbf{j}} $$
So, magnitude of acceleration at $t=1 s$ is
$$ \begin{gathered} |\mathbf{a}| _{t=1 s}=\sqrt{a _x^{2}+a _y^{2}}=\sqrt{30^{2}+40^{2}} \\ =50 ms^{-2} \end{gathered} $$
BETA
