Question

The acceleration of an object is given by $a(t) = \cos \pi t m{s^{ - 2}}$ and its velocity at time $t = 0$ is$\dfrac{1}{{2\pi }} m{s^{ - 1}}$ at origin. Find its velocity at time $t = \dfrac{3}{2}s$.
A. $ - \dfrac{1}{{2\pi }}m{s^{ - 1}}$
B. $\dfrac{3}{{2\pi }}m{s^{ - 1}}$
C. $\dfrac{1}{\pi }m{s^{ - 1}}$
D. $\dfrac{2}{\pi }m{s^{ - 1}}$

Answer 1

Hint: Integrate acceleration w.r.t. time and obtain the velocity of the object as a function of time. Then, substitute $t = \dfrac{3}{2}s$ to the equation to calculate the velocity at 3/2 seconds.

Complete step by step answer:

We know that, since acceleration is defined as the rate of change of velocity, velocity can be expressed as the integration of acceleration with respect to time. We have,
$ \Rightarrow v(t) = \int {a(t).dt} $
Given the value of acceleration from the question, we get,
$ \Rightarrow v(t) = \int {\cos \pi t.dt} $
To solve the integral, we substitute, $\pi t = u,$ then $du = \pi dt$ in the above equation,
$ \Rightarrow v(u) = \int {\cos u.\dfrac{{du}}{\pi }} $
Since $\pi $ is a constant, we can take $\dfrac{1}{\pi }$ out of the integral, we get,
$ \Rightarrow v(u) = \dfrac{1}{\pi }\int {\cos u.du} $
Now we know that $\int {\cos (u).du = \sin (u) + C} $, hence we get-
$ \Rightarrow v(u) = \dfrac{1}{\pi }\sin (u) + C$
Substituting back $u = \pi t,$ we get,
$ \Rightarrow v(t) = \dfrac{1}{\pi }\sin (\pi t) + C$ .................. Equation 1
Given at time $t = 0,v = \dfrac{1}{{2\pi }}m{s^{ - 1}}$ substituting in the above equation, we get
$ \Rightarrow \dfrac{1}{\pi }\sin (\pi \times 0) + C = \dfrac{1}{{2\pi }}$
Since $\sin {0^ \circ } = 0$ , we get,
$ \Rightarrow C = \dfrac{1}{{2\pi }}$
Putting the value of C above in equation 1 we get,
$ \Rightarrow v(t) = \dfrac{1}{\pi }\sin (\pi t) + \dfrac{1}{{2\pi }}$ .................. Equation 2
Now, at velocity at $t = \dfrac{3}{2}s$ is given by:
$ \Rightarrow v = \dfrac{1}{\pi }\sin \left( {\dfrac{{3\pi }}{2}} \right) + \dfrac{1}{{2\pi }}$
Putting the value of $\sin \left( {\dfrac{{3\pi }}{2}} \right) = - 1$ in the above equation, we get,
$ \Rightarrow v = - \dfrac{1}{\pi } + \dfrac{1}{{2\pi }} = - \dfrac{1}{{2\pi }}m{s^{ - 1}}$
Hence, at time$t = \dfrac{3}{2}s$, we get the velocity as $ - \dfrac{1}{{2\pi }}m{s^{ - 1}}$.
Therefore the correct option is A.

Note: Alternate Method, instead of solving the indefinite integral and obtaining equation 1 and then finding and substituting the value of C, we can also use the definite integral with limits as follows,
We know that,
$ \Rightarrow \dfrac{{dv}}{{dt}} = a(t)$
Bringing dt to the other side of the equation, we get
$ \Rightarrow dv = a(t).dt$
Integrating both sides with limits for t as 0 to t and for corresponding limits for velocity we use $v =\dfrac{1}{{2\pi }}$ when t=0; we get
$ \Rightarrow \int\limits_{\dfrac{1}{{2\pi }}}^v {dv = \int\limits_0^t {a(t).dt} } $
And substitute the equation given for acceleration, we get
$ \Rightarrow \int\limits_{\dfrac{1}{{2\pi }}}^v {dv = \int\limits_0^t {\cos (\pi t).dt} } $
Solving the definite integral is identical to the method in the answer and we get
$ \Rightarrow \left[ v \right]_{\dfrac{1}{{2\pi }}}^v = \left[ {\dfrac{1}{\pi }\sin \left( {\pi t} \right)}
\right]_0^t$
Substituting the limits, we obtain Equation 2
$ \Rightarrow v - \dfrac{1}{{2\pi }} = \dfrac{1}{\pi }\sin (\pi t) - 0$
And the rest of the steps are the same as given in the step-by-step answer after Equation 2.