Question

If velocity of a particle depends on time (t) as v= (4cos t + t²) m/s then displacement (in m) of the particle from t=0 t=π/2 s will be
a) π+π³/24
b) π³/24
c) 4+π³/24
d) π​

Answer 1

Explanation:☝️☝️

Answer:

$4 + \\pi}^{3}} \div 24$

Answer 2

Given,

Velocity of a particle depends on time (t) as v = (4cos t + t²) m/s

To find,

Displacement (in m) of the particle from t=0 t=π/2 s

Solution,

The velocity of a moving object such as a particle is defined as the rate of change of its displacement with respect to time. It is given as,

$v=\frac{ds}{dt}$

Where,

v = velocity

s = displacement, and,

t = time.

Rearranging this equation, we get,

$ds =v\cdot dt$

Now, the displacement from time $t_1$ to $t_2$ can be found by integrating the above relation as,

$ds=\int\limits^{t_2}_{t_1} {v} \, dt$

Here, it is given that,

v = (4cos t + t²), thus,

$ds=\int\limits^{\frac{\pi }{2} }_0 {(4\cos t + t^2)} \, dt$

$\implies ds=4\int\limits^{\frac{\pi }{2} }_0 {(\cos t)} \, dt+\int\limits^{\frac{\pi }{2} }_0 {(t^2)} \, dt$

$\implies s=4[\sin t]\Biggr|_{0}^{\frac{\pi }{2} }+[\frac{t^3}{3}] \Biggr|_{0}^{\frac{\pi }{2} }$

$\implies s=4[\sin t]\Biggr|_{0}^{\frac{\pi }{2} }+\frac{1}{3}[t^3] \Biggr|_{0}^{\frac{\pi }{2} }$

$\implies s=4[\sin \frac{\pi }{2} -\sin0]+\frac{1}{3}[(\frac{\pi }{2} )^3-0^3]$

$\implies s = 4[1 - 0] + \frac{1}{3} \cdot \frac{\pi ^3}{8}$

$\implies s=4+\frac{\pi ^3}{24}$ m, which is the required displacement.

Therefore, the displacement of the particle (in m) will be $4+\frac{\pi ^3}{24}$. (option c).