Question

A mosquito is moving with a velocity \[\overrightarrow v = 0.52{t^2} \hat {i}+ 3t \hat {j} + 9 \hat {k}\]m/s and accelerating in uniform conditions. What will be the direction of the mosquito after \[2s\]?
a. ${\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$from x axis
b. ${\tan ^{ - 1}}\left( {\frac{5}{2}} \right)$from x axis
c. ${\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$from y axis
D. ${\tan ^{ - 1}}\left( {\frac{5}{2}} \right)$from y axis

Answer 1

Hint: velocity per unit time is defined as acceleration. Thus, differential of velocity with respect to time gives acceleration.
In trigonometry, \[tan{\text{ }}x\] is a function given by ratio of perpendicular of the triangle having x as its base angle to the base of the triangle.

Complete answer:
Step $1$: a mosquito is moving with velocity having equation:
\[\overrightarrow v = 0.52{t^2} \hat {i} + 3t\hat {j} + 9\hat {k}\]
Thus, differentiating velocity with respect to time to find acceleration vector
$\overrightarrow a = \frac{{d\overrightarrow v }}{{dt}}$
\[\overrightarrow a = 1.04t\hat {i} + 3\hat {j} \]

Step $2$: Given the value of time is $2$ seconds
Thus, \[\overrightarrow a = 1.04(2)\hat {i} + 3\hat {j}\]
Or \[\overrightarrow a = 2.08\hat {i} + 3\hat {j}\]
Thus, from the above acceleration vector it is clear that the value of x component is $2.08$ or approximately $2$ and the value of y component is $3$. As the x,y components are $2,3$ respectively so option b and d are ruled out as 5 and 2 cannot be the direction.

Step $3$: Now the aim is to find out whether the angle whose value is ${\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$ is on x axis or y axis.
We know that, \[tan\theta {\text{ }}\]$ = \frac{y}{x}$if $\theta $ is the angle with the x axis.

Let us assume $\theta $ to be on x axis.

Thus, $\theta $ is ratio of y component to x component so $\theta = {\tan ^{ - 1}}\left( {\frac{3}{2}} \right)$. This means the direction of mosquito will be ${\tan ^{ - 1}}\left( {\frac{3}{2}} \right)$ from x axis. But there is no option with this answer.

Hence, changing our approach to y axis.
Now $\theta $ is on y axis. So, \[tan\theta \]$ = \frac{x}{y}$
Thus, $\theta $ is ratio of x component to y component so $\theta $$ = {\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$.

Hence, the direction of mosquito after $2$ sec will be ${\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$from y axis. Thus, correct option is c.

Note: i cap signifies direction of x axis and j cap signifies direction of y axis and k cap signifies direction of z axis. As the differentiation of velocity vector ruled out the k cap component because it was a constant quantity. This clearly implies that the acceleration of mosquito is restricted to two dimensions.