Question

For a freely falling object, a(t) = - 32 ft/sec^2, v(0) = initial velocity = v0 (in ft/sec), and s(0) = initial height = S0 (in ft). Find a general expression for s(t) in terms of v0 and s0.

Answer 1

Answer:

$s(t)=s_{0}+v_{0}t-(0.5)(32)t^{2}$

Step-by-step explanation:

Let's use the definition of acceleration.

$a(t)=dv/dt$

If we take the integral in both sides we will have:

$\int\limits^t_{t_{0}} {a(t)} \, dt=\int\limits^v_{v_{0}} {dv}$

a(t) = -32, so it is independent of time.

$a(t)\int\limits^t_{t_{0}} {dt}=\int\limits^v_{v_{0}} {dv}$

$a(t)(t-t_{0})=v-v_{0}$

we can assume that $t_{0} = 0$

$v(t)=v_{0}+a(t)t$ (1)

Using the definition of v(t) as the derivative of s (height) with t (time) we have:

$v=ds/dt$(2)  

Taking the integral in both sides we can find s(t), and using (1) we have:

$\int\limits^t_{t_{0}} {v(t)} \, dt=\int\limits^s_{s_{0}} {dx}$

Using (1) in (2)

$\int\limits^t_{t_{0}}( v_{0}+ta(t))\, dt=\int\limits^s_{s_{0}} {ds}$

solving this integral, we have:

$v_{0}t+0.5a(t)t^{2}=s(t)-s_{0}$

Finally, let's solve this equation for s(t).

$s(t)=s_{0}+v_{0}t+0.5a(t)t^{2}$

$s(t)=s_{0}+v_{0}t-(0.5)(32)t^{2}$

Have a nice day!