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lara [203]
3 years ago
9

The position function of an object moving along a straight line is given by s=f(t). The average velocity of an object over the t

ime interval [a,b] is the average rate of change of f over [a,b]; it’s velocity at t=a is the rate of change of f at a.
A ball is thrown straight up with an initial velocity of 144ft/sec, so that it’s height (in feet) after t sec is given by s=f(t)= 144t-16t^2.
(A)What is the average velocity of the ball over the following time intervals?
[3,4]= ft/sec
[3,3.5]= ft/sec
[3,3.1]= ft/sec

(B) what is the instantaneous velocity at time t=3? Ft/sec

(C) what is the instantaneous velocity at time t=8? Ft/sec

(D) when will the ball hit the ground?
Mathematics
1 answer:
Oksana_A [137]3 years ago
3 0
A) Position at time t = 3 s : 
<span>s₁ = 144×3 - 16×3² = 288 ft </span>
Position at time t = 4 s : 
<span>s₂ = 144×4 - 16×4² = 320 ft 
</span>
<span>Δs = s₂ - s₁ = 320 - 288 =  32ft </span>
<span>Δt = t₂ - t₁ = 4 - 3 = 1 s 

</span><span>Vavg[3, 4] = Δs / Δt = 32 / 1 = <u>32 ft/s</u>
</span><span>Vavg[3 , 3.5] = [(144×3.5 - 16×3.5²) - (144×3 - 16×3²)] / [3.5 - 3] = <u>40 ft/s</u>
</span><span>Vavg[4 , 4.1] = [(144×3.1 - 16×3.1²) - (144×3 - 16×3²)] / [3.1 - 3] = <u>4.64 ft/s</u></span>

B) lim [(144t - 16t²) - (144×4 - 16×3²)) / (t - 3)] = 
<span>t→3
</span>lim [(-16t² + 144t - 288) / (t - 3)] = 
<span>t→3
</span>
<span><u>48 ft/s</u>

</span>C) <span>lim [(144t - 16t²) - (144×8 - 16×8²)) / (t - 8)] = </span>
<span>t→8 </span>

<span>lim [(-16t² + 144t - 128) / (t - 8)] = </span>
<span>t→8 </span>

<span>lim (-16t + 16) = </span>
<span>t→8 </span>

<span><u>-112 ft/s</u> 

D) </span><span>144t - 16t² = 0 </span>
<span>(144 - 16t) t = 0 </span>
<span>t = 0 </span>
<span>or </span>
<span>144 - 16t = 0 
</span>
<span><u>t = 9 </u></span>
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