an object is dropped from a tower, 196 ft above the ground. the objects height above ground t sec into the…
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} ).\na. what is the objects velocity, speed, and acceleration at time t?\nb. about how long does it take the object to hit the ground?\nc. what is the objects velocity at the moment of impact?\nthe objects velocity at time t is
Answer
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.