velocity of a car using by evalulating definite integral?

Question

a car moves along a straight road in such a way that its velocity at any time in t seconds is given by

v(t) = 3t√16-t^2
0≤t≤4
find the distance traveled by the car in 4 from t=0 to t=4

I have no idea what to do. In this chapter they list the properties of a definite integral and the average value of a function. I don't know what tool to use.

Answer 1

Integrating the velocity function from 0 to 4 will give you the distance the car travels.

3 *Integ[ (16 - t²) ^(1/2) t dt ]
-3/2 *Integ[ (16 - t²) ^(1/2) * -2t dt]
-3/2 * [2/3 (16 - t²) ^(3/2) | 4 -> 0
-3/2 [ 0 - 128/3]
= 64 feet

Hope this helps :D

Answer 2

If you are moving at constant velocity then the distance traveled is just velocity time the time or V*T. But if the velocity is not constant with time, as in this problem, then the velocity must be integrated with time to get the distance.

Distance = INTEGRAL (v(t) dt) evaluated from t = 0 to 4

INTEGRAL (3t SQRT(16 - t^2) dt)

Use U = 16 - t^2 and dU = -2t
And limits of integration become U=16-0^2=16 and U=16-4^2=0

INTEGRAL (-(3/2) SQRT(U)) dU) evaluated from U=16 to 0
the INTEGRAL( SQRT(U) dU) = [ U^(3/2) ]/(3/2)
(3/2) /(3/2) = 1 so the Integral is just:
-U^(3/2) = -(0)^(3/2) - [-(16)^(3/2)] = 64

Answer 3

s=∫3t√16-t^2 dt from0 to 4 , put 16-t^2= u^2

then -2tdt =2udu s=∫-3u^2 du from 4 to 0
s= - u^3 from 4 to 0 then s = 0 -(- 4)^3 = 64 units of length