Question

An angle $\theta$ through which a pulley turns with time ‘t’ is completed by $\theta ={{t}^{2}}+3t-5$ sq.cms/min. Then the angular velocity for t = 5 sec.
(a) ${{5}^{c}}/\sec$
(b) ${{13}^{c}}/\sec$
(c) ${{23}^{c}}/\sec$
(d) ${{35}^{c}}/\sec$

Answer 1

Hint: Firstly calculate the value of angular velocity v in terms of an expression by differentiating the angular position $\theta$ with respect of time t. Then substitute the value of time t = 5 sec in the expression of angular velocity v and calculate the final answer.

Complete step-by-step answer:
We know that angular velocity v is the rate of change of angular position $\theta$ with respect to time t.
The above statement can be expressed mathematically as,
$v=\dfrac{d\theta }{dt}$
It is said that $\theta ={{t}^{2}}+3t-5$.
By substituting the value of angular position $\theta$ in the above expression we get,
$v=\dfrac{d}{dt}\left( {{t}^{2}}+3t-5 \right)$
Differentiating the above expression we get,
$v=2t+3$
Now, we have to find the angular velocity for t = 5 sec, therefore substituting t = 5 in the above expression,
$v=2\left( 5 \right)+3$
Multiplying 2 with 5 and simplify,
$v=10+3$
Adding 3 to 10 we get,
v = 13
We conclude that the angular velocity v at t = 5 sec is ${{13}^{c}}/\sec$.
Hence option (b) is the correct answer.

Note: The possible mistake that you may encounter could be that the differentiation might not be correct. It should be taken into consideration that the differentiation being done above is in terms of time t and not any other variable x or y. Substitution of values should be done correctly.