Integral Problem!!?

Question

Having problem with an intergral problem!!!?
I have the equation v(t) = 5t mph, [t is in seconds] and I have to find the total distance in miles in its accelaration. And the car accelerates from 0 to 25 mph.. I think the seconds conversion is messing with my brain.


oh and I posted this by accident in the History section... oops. I fail

Answer 1

Express v(t) in miles per second, and then integrate it from 0 to 5, since v(5) = 25 mph.

25 mph = 25 miles / 1 hour * 1 hour / 3600 sec = 25/3600 mps.

Acceleration is constant, so v(t) = ct for some c. Since you want v(5) = 5c = 25/3600, it follows that c = 5/3600.

Therefore, to get the distance, you need to compute the integral from 0 to 5 of 5t/3600 dt.

Answer 2

v(5) = 25 mph.

25 mph = (25 miles/hour)(1 hr/3600 s) = 25/3600 mps.

v(5) = 5t/3600

Now, you just need to integrate the above expression from 0 to 5.

∫5t/3600 dt (from 0 to 5)

= (5/2)(25)^2/3600

= .43 miles

Answer 3

V(t)=5t mph=5t/3600 mps
a=dV(t)/dt=5 mph every second=5/3600 mps^2
For the car to reach max. speed of 25 mph,
V(t)=a*t=25
t=25/a
t=25/5=5 seconds
S(t)=integral (V(t).dt over 0<=t<=5) [S= distance]
S(t)=integral (5t/3600.dt over 0<=t<=5)
S(t)=(5/2)t^2/3600 over 0<=t<=5
S(t)=2.5*(5^2-0)/3600
S(t)=62.5/3600 miles
S(t)=0.017 miles
s(t)=0.017 miles
or, S(t)=1/2*a*t^2
S(t)=1/2*5/3600*5^2
S(t)=0.017 miles