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If an object is dropped from a tall buil...

If an object is dropped from a tall building , then the distance it has fallen after t seconds is given by `d(t)=16t^2` . Find its average speed , in feet per second, between t=1 second and t=5 seconds.

A

198

B

192

C

96

D

80

Text Solution

AI Generated Solution

The correct Answer is:
To find the average speed of an object dropped from a tall building between \( t = 1 \) second and \( t = 5 \) seconds, we will follow these steps: ### Step 1: Understand the distance function The distance fallen after \( t \) seconds is given by the function: \[ d(t) = 16t^2 \] ### Step 2: Calculate the distance at \( t = 1 \) second Substituting \( t = 1 \) into the distance function: \[ d(1) = 16(1^2) = 16 \text{ feet} \] ### Step 3: Calculate the distance at \( t = 5 \) seconds Substituting \( t = 5 \) into the distance function: \[ d(5) = 16(5^2) = 16(25) = 400 \text{ feet} \] ### Step 4: Find the total distance fallen between \( t = 1 \) and \( t = 5 \) The total distance fallen from \( t = 1 \) to \( t = 5 \) is: \[ \text{Total distance} = d(5) - d(1) = 400 - 16 = 384 \text{ feet} \] ### Step 5: Calculate the time interval The time interval between \( t = 1 \) and \( t = 5 \) seconds is: \[ \Delta t = 5 - 1 = 4 \text{ seconds} \] ### Step 6: Calculate the average speed The average speed \( v_{\text{avg}} \) is given by the formula: \[ v_{\text{avg}} = \frac{\text{Total distance}}{\Delta t} \] Substituting the values we found: \[ v_{\text{avg}} = \frac{384 \text{ feet}}{4 \text{ seconds}} = 96 \text{ feet per second} \] ### Final Answer The average speed of the object between \( t = 1 \) second and \( t = 5 \) seconds is: \[ \boxed{96 \text{ feet per second}} \]
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