Question

an object is dropped from a tower, 196 ft above the ground. the objects height above ground t sec into the fall is ( s = 196 - 16t^{2} ).
a. what is the objects velocity, speed, and acceleration at time t?
b. about how long does it take the object to hit the ground?
c. what is the objects velocity at the moment of impact?
the objects velocity at time t is

Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s(t)\). Given \(s(t)=196 - 16t^{2}\), using the power rule \(\frac{d}{dt}(x^{n})=nx^{n - 1}\), we have \(v(t)=\frac{ds}{dt}=\frac{d}{dt}(196)-\frac{d}{dt}(16t^{2})\).
Since \(\frac{d}{dt}(c) = 0\) (where \(c\) is a constant) and \(\frac{d}{dt}(16t^{2})=16\times2t\), then \(v(t)=- 32t\).

Step2: Find the speed function

Speed is the absolute - value of velocity. So, speed \(=\vert v(t)\vert=\vert-32t\vert = 32t\) (since \(t\geq0\) in the context of time for the fall).

Step3: Find the acceleration function

The acceleration \(a(t)\) is the derivative of the velocity function. Given \(v(t)=-32t\), using the power rule \(\frac{d}{dt}(x^{n})=nx^{n - 1}\) with \(n = 1\), we have \(a(t)=\frac{dv}{dt}=-32\).

Answer:

The object's velocity at time \(t\) is \(-32t\) ft/sec, the speed is \(32t\) ft/sec, and the acceleration is \(-32\) ft/sec².