What is the average speed of f(t)= t^2+t?

Question

Between t=1 and t=2

Answer 1

Average velocity = (change in x) / (change in time)

Average velocity = [f(2) - f(1)] / (2 - 1)

Speed is the absolute value of velocity:

f(2) = 2^2 + 2

f(2) = 4 + 2

f(2) = 6

f(1) = 1^2 + 1

f(1) = 1 + 1

f(1) = 2

Avg. speed = (6 - 2) / (2 - 1)

Avg. speed = 4/1

Avg. speed = 4 units/second

Answer 2

dy/dt=2t+1

= 3 (t=1)

=5(t=3)

Avg =4  

Answer 3

Average speed is calculated like slope.

f(2)-f(1) / 2-1

6 - 2 = 4



4/1 =4

Answer 4

f(t) = t² + t

f(1) = 2

f(2) = 5

Average = (5-2) / 1 = 3

Assuming f(t) = distance

.

Answer 5

I don't know what f(t) is.

Assuming f(t) is x(t), the coordinate of the moving point at instant t

x(t) = t^2 + t

x(1) = 1^2 + 1 = 2

x(2) = 2^2 + 2 = 6

then

v(average) = Δx / Δt = [x(2) - x(1)] / (2 - 1) = 4

Answer 6

If y(f(t)) represents displacement, then average speed is 3 m/s in the 2nd second...If y is speed itself then the answer is same...