Question

If 𝑎 = 4𝑡^−3/2 , 𝑠 = 16 𝑤ℎ𝑒𝑛 𝑡 = 4, 𝑎𝑛𝑑 𝑠 = 25 𝑤ℎ𝑒𝑛 𝑡 = 6, find the equation of motion 𝑠 = 𝑓(𝑡) and the velocity function 𝑣(𝑡).

Answer 1

Acceleration $a$ is given as

$$a = \dfrac{dv}{dt} \implies dv = a\;dt \implies v = \int a\;dt$$

The function of the velocity ($v$) is given as:

$$v =\int 4t^{-3/2} dt\\ = 4 \int t^{-3/2} dt\\ = 4\int \dfrac{1}{t^\frac{3}{2}} dt\\ = 4 \Bigg(-\dfrac{2}{\sqrt{t}}\Bigg)\\ v=-\dfrac{8}{\sqrt{t}} + c$$

Velocity ($v$) is given as:

$$v = \dfrac{ds}{dt} \implies ds = v\; dt \implies s = \int v\;dt$$

The function of the equation of motion can be obtained as follows:

$$s= \int \Big(-\dfrac{8}{\sqrt{t}} + c \Big) dt\\ = -8 (2\sqrt{t}) + ct\\ s = - 16\sqrt{t} + ct +c_1$$

When s = 16 and t = 4:

$$16 = -16(\sqrt{4})+c(4) + c_1\\ = -16(2) + 4c + c_1\\ 16= -32 +4c +c_1\\ \implies 4c+c_1 = 48 \qquad\dots (eqn\;1)$$

When s = 25 and t = 6:

$$25 = -16(\sqrt{6})+c(6) + c_1\\ = -16(\sqrt{6}) + 6c + c_1\\ 25= -16(2.45) +6c +c_1\\ 6c+c_1 = 25+39.2 \\ \implies 6c+c_1 = 64.2 \qquad\dots (eqn\;2)$$

Solving eqn 1 and eqn 2 simultaneously:

$$\quad4c+c_1=48\\ (-) 6c+c1=64.2\\ ---------\\ -2c=-16.2\\ c= 8.1$$

Hence we find $c_1$ :

$$4(8.1) +c_1 = 48\\ 32.4 +c_1 = 48\\ c_1 = 48 -32.4\\ c_1=15.6$$

Therefore, the equation of the motion $s = f(t)$ is:

$$s=f(t)=-16\sqrt{t}+8.1t+15.6$$

The velocity $v(t)$ is:

$$v(t) = -\dfrac{8}{\sqrt{t}} + 8.1$$