Kinematics 5 Question 3

5. 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: 5. (a)

Solution:

  1. 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

$$ \begin{gathered} |\mathbf{a}| _{t=1 s}=\sqrt{a _x^{2}+a _y^{2}}=\sqrt{30^{2}+40^{2}} \\ =50 ms^{-2} \end{gathered} $$



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