basic integrals

Question

A stone is thrown from the top of a tall cliff. Its acceleration is a constant​ (-32)

ft/sec2  ​(So A(t)​ = -32). Its velocity after 2 seconds is

(27) ft/sec​, and its height after 2

seconds is 214 ft.

Find the velocity function.

​v(t) = enter your response here

Find the height function.

​h(t) = enter your response here

Answer 1

Assuming the object moves in the vertical direction

we apply the standard formulas for motion with an uniform acceleration a=-32ft/s^2

v(t)=v(0)+at (1)

h(t)=h(0)+v(0)t +(at^2)/2 (2)

We have just two unknows here: v(0) and h(0)

1. v(0) can be found from (1) using v(2)=27ft/s :

v(0)=v(2)-2a=27ft/s-((-32 ft/s^2)(2s))=27ft/s+64ft/s=91ft/s (3)

2. Find h(0)

Substituting v(0) from (3) into (2) at t=2s, we'd solve for h(0) (using the given h(2)=214ft )

:

h(0)=214ft -2sx91ft/s -((-32ft/s^2)x (2s)^2)/2=214ft-182ft+64ft=96ft

So the result can be written as (all lengths are in ft, time in s)

v(t)=91-(32) t

h(t)=96+(91) t -(32t^2)/2

Answer 2

Hi Alecia,

a(t)=-32

velocity is the integral (antiderivative) of acceleration.

v(t)=-32t+C (don't forget the plus C!)

It's given that v(2)=27------> 27=32(2)+C --------> 27=64+C---------> C=27-64=-37

So v(t)=-32t-37

Your displacement, h(t) is the antiderivative of velocity. Reverse power rule you get:

h(t)=-16t^2-37t+C

It's given that h(2)=214 ------> 214=-16(2)^2-37(2)+C----------> 214=-64-74+C ---------->C=214+64+74=352

so h(t)=-16t^2-37t+352