Question

the velocity time relationship of a body starting from rest is given by v=k^2t^3/2,where k is √2m^1/2 /s^5/2,Find the distance traversed by the body in 4 seconds

Answer 1

the velocity time relationship of a body starting from rest is given by $v=k^2t^{3/2}$, here $k=\frac{\sqrt{2}m^{1/2}}{s^{5/2}}$

Let's put value of k in equation ,
$v=2ms^{-5}t^{3/2}\\\\\frac{dx}{dt}=2ms^{-5}t^{3/2}\\\\\int{dx}=\int{2ms^{-5}t^{3/2}dt}$
$\int\limits^x_0{dx}=\int\limits^4_0{2ms^{-5}t^{3/2}dt}$
$x=\left[\begin{array}{c}\frac{2mt^{5/2}}{\frac{5}{2}s^5}\end{array}\right]^4_0$
$x=\frac{4m}{5s^5}(2)^5$

Answer 2

Answer:

hey, see the solution attached.