Suppose an object is dropped from a lunar cliff and its speed in ft/sec. after t seconds is given by v(t)=5.3t

Question

If the object lands after 5 seconds, how high (in ft) is the cliff?

Answer 1

Since the function is linear, calculus is not really necessary.

The velocity function is v(t) = 5.3t, so after 5 seconds it's dropping at 5.3(5), or 26.5 ft/sec

Picture a graph in the form of a right triangle with time on x-axis and velocity on y-axis. At t = 0, velocity is 0.

The max height of the triangle after 5 seconds is 26.5, the base is 5 sec.

So the area of the triangle (represents distance) will be 26.5(5)/2 = 132.5/2 = 66.25 ft.

Answer 2

d(t)=∫v(t)dt is a physics relation,,,d=distance

here d(5)-d(0)=∫5.3t*dt evaluated t=0 to t=5

d=5.3*25/2=66.25 ft.

Answer 3

Plug in t=5:

v(t) = 5.3t

v(5) = 5.3(5) = 26.5

Thus, the cliff is 26.5ft high.

Hope this helped!