I need help with calculus problem

as you drive along the highway,you step hard on the accelerator to pass a truck. Assume that your velocity, v(t) feet per second is given by v(t)=40+5√t, where t is the number of seconds since you started accelerating. Find an equation for D(t), your displacement from starting point, that is, from D(0)=0. How far do you go in the 10 sec it takes to pass the truck
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Start by rewriting the velocity equation in terms of powers, which yields:

v(t) = 40 + 5t^(1/2)

Understand that the velocity function is actually the first derivative of the displacement function...in other words...v(t) = D ' (t). Since we want D(t), we need to integrate.

Integral of 40 + 5t^(1/2) --> Integrate

D(t) = 40t + (10/3)t^(3/2) + C

Use D(0) = 0 to solve for C, which yields:

0 = 40(0) + (10/3)(0)^(3/2) + C --> Simplify.

C = 0

Therefore our final equation for D(t) is:

ANSWER: D(t) = 40t + (10/3)t^(3/2)

To answer the second part, "How far do you go in 10 seconds", simply substitute t = 10 and solve for the distance...

D(10) = 40(10) + (10/3)10^(3/2) --> Simplify.

D(10) = 400 + (10/3)[ 10Sqrt(10) ] --> Use a calculator to approximate.

ANSWER: D(10) = 505 feet

Hope that helps!

to get D(t) take the integral of v(t) equation. Which is V(t)= 40t+10/3(t)^(3/2) + c. To find out what c is put in 0 for t since D(0)=0, therefore c=0 and D(t)=40t+10/3(t)^(3/2), put in 10 seconds for t and you get 505.4 ft