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I have the following question:

An object is dropped from a tower $184$ft above the ground. The object's height above the ground $t$ sec into the fall is $s=184-16t^2$

1) What is the object's velocity at time $t$? (I got $-32t$)

2) What is the object's speed at time $t$? (the answer they are looking for is in $\frac{\text{ft}}{\text{sec}^2}$)

Is the answer $32t \frac{\text{ft}}{\text{sec}^2}$?

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    wouldn't speed be 32 $\frac{ft}{sec}$ and acceleration be $32\frac{ft}{sec^2}$? I am assuming that because the object is falling downward and picking up speed then its acceleration must be positive, but the velocity is $-32t$ wouldn't that require acceleration to be $-32$ because it is the derivative of velocity? – 2017-02-18
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    In general, velocity is a vector and speed is the magnitude of that vector. In your example, the velocity is $-32t\frac{ft}{sec}$; indicating that the speed is $32t\frac{ft}{sec}$ and the direction is "down" (assuming that $t \ge 0$). – 2017-02-18

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