object is thrown down from a cliff with an initial speed of 4 feet per sec, and its speed in ft/sec after t seconds is given by s(t)=32t+4.

Question

If the object lands after 8 seconds, how high (in ft) is the cliff?

Answer 1

As a linear function, it can be done without calculus.

v(t) = 32t + 4

If you graph it on an xy coordinate plane with time on the x-axis and velocity on the y-axis, you'll get a straight line from t = 0 (y = 4) to 32(8) + 4 = 260 ft/sec after t = 8 secs.

The area under the line represents the product of rate x time, or distance.

You can break the area down into a rectangle with dimensions 4 x 8 = 32

and a right triangle with area base x height divided by 2, or 8(260 - 4)/2 = 1024

Adding the two parts yields a distance of 1056 ft.

Answer 2

normally speed is v(t) not s(t) just to not

here distance(t) = d(t)=∫(32t+4)dt evaluated from t=0 to t=8

d=16t²+4t evaluated t=0 to t=8

d=1056 ft.

Answer 3

Please see my previous answer