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Olenka [21]
3 years ago
14

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.
Mathematics
1 answer:
vlada-n [284]3 years ago
3 0

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

<u>If we take the integral in both sides we will have:</u>

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

<em>a(t) = -32, so it is independent of time.</em>

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

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

<em>we can assume that t_{0} = 0</em>

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)  

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

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

<u>Using (1) in (2)</u>

\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!

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Hello there! Thank you for asking your question here at Brainly! I will be assisting you today with answering this problem, and will teach you how to handle it on your own in the future.

First, let's take a look at our problem.
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