Find the car's average velocity

Question

A car drives down a road in such a way that its velocity (in m/s) at time t (seconds) is

v(t)=2t^1/2+5

Find the car's average velocity (in m/s) between t = 2 and t = 6.

Answer 1

To find the average, we find area under the curve between t=2 and t=6.  We have

 

⁶₂ ∫(2t^1/2 + 5)dt

 

We can break up the integral as

 

 2 ⁶₂∫t^1/2 dt  +  5 ⁶₂∫dt  =

 

= [2(2/3)t^3/2 + 5t] |⁶₂ 

 

= (4/3)t^3/2 + 5t) |⁶₂  

 

= [(4/3)(6)^3/2 + 5(6)] - [(4/3)(2)^3/2 + 5(2)]

 

= [(4/3)14.70 + 30]  -  [(4/3)2.83 + 10]

 

= 49.6  -  13.78

= 35.82

 

The cars average velocity is 35.3 m/s²

 

Answer 2

We want to find the average velocity. You'll want to integrate v(t) from your two listed t values (from 2 and 6). Then you divide by t=(t2-t1)=4 s. 

 

Your integral starts off with your given equation.

 

⁶∫₂ 2t^(1/2)+5 dt

= (4/3)t^(3/2)+4t

 

Plug in the value of 6 and plug in the value of 2. subtract the two and then divide by 4 (because remember, this is the average for velocity).