Question

The velocity time relation of a body starting from rest is given by v=k2t3/2 where k=√2m1/2/s5/4 the distance traversed in 4seconds is

Answer 1

Answer:

hey,see the attached solution.

Answer 2

The velocity - time relation of a body starting from rest is given by, $v=k^2t^{3/2}$, where k = $\sqrt{2}m^{1/2}s^{5/4}$

so, we have to find distance travelled in 4 seconds.

we know, velocity is the rate of change of displacement with respect to time.

i.e., dx/dt = v

or, $\int{dx}=x =\int\limits^t_0{v}\,dt$

so, distance travelled in 4 seconds, s = $\int\limits^4_0{k^2t^{3/2}}\,dt$

= $k^2\left[\frac{k^{5/2}}{5/2}\right]^4_0$

= $\frac{2k^2}{5}[(4)^{5/2}]$

= $\frac{2k^2}{5}[2^5]$

= $\frac{64k^2}{5}$

putting value of k = $\sqrt{2}m^{1/2}s^{5/4}$

so, s = 64(√2)²/5 = 128/5

hence, distance travelled in 4 seconds = 128/5 m