V(t) = t^(1/2) + 1 = sqrt(t) + 1.
V(2) = 2^(1/2) + 1 + 2.41 m/s
V(6)= 6^(1/2) + 1 = 3.45 M/S
Vavg = (2.41 + 3.45) / 2 = 2.93 m/s
A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is
v(t)=1t^(1/2)+1
.
Find the car's average velocity (in m/s) between t=2 and t=6. ?
I am lost!
2 answers
If we consider the average velocity as the total distance travelled in a given time, then
V̄
= [∫(sqrt(t)+1)dt ] / (6-2)
= [(2/3)t^(3/2)+t] from 2 to 6 / (6-2)
= (15.80/-3.88)/4
=2.98 m/s
V̄
= [∫(sqrt(t)+1)dt ] / (6-2)
= [(2/3)t^(3/2)+t] from 2 to 6 / (6-2)
= (15.80/-3.88)/4
=2.98 m/s
1 answer