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 +2 .
Find the car's average velocity (in m/s) between t=4 and t=9.
i really need help with thisss
To find the car's average velocity between t=4 and t=9, you need to calculate the total displacement and divide it by the time interval.
The velocity function given is v(t) = 2t^(1/2) + 2.
To find the displacement, you need to integrate the velocity function over the time interval [4, 9]:
∫[4,9] (2t^(1/2) + 2) dt
Integrating the function 2t^(1/2) + 2 with respect to t gives:
(2/3)t^(3/2) + 2t
Now, you can evaluate the definite integral over the interval [4, 9]:
[(2/3)(9)^(3/2) + 2(9)] - [(2/3)(4)^(3/2) + 2(4)]
Simplifying further:
[(2/3)(27) + 18] - [(2/3)(8) + 8]
[(54/3) + 18] - [(16/3) + 8]
(54/3 + 54/3) - (16/3 + 24/3)
108/3 - 40/3
68/3
Therefore, the total displacement over the time interval [4, 9] is 68/3 m.
Now, you need to divide the displacement by the time interval to find the average velocity:
Average velocity = Displacement / Time interval
= (68/3) / (9 - 4)
= (68/3) / 5
= 68/15
= 4.53 m/s (rounded to two decimal places)
So, the car's average velocity between t=4 and t=9 is 4.53 m/s.
V(4) = 2 * sqrt(4) + 2 = 4 + 2 = 6m/s.
V(9) = 2 * sqrt(9) + 2 = 6 + 2 = 8.
V(avg.) = (6 + 8) / 2 = 7m/s.