The speed of a car, v kilometres per hour... word problem

[Me_self]

the speed of a car, v kilometres per hour, at a time of t hours is represented by v(t) =40 +3t -t^2. the rate of gazoline consumtion of the car, c litres per kilometre, at a speed of v kilometres per hour is represented by c(v) =(v/500-0.1)^2 +0.15. determine an algebraic expression for c(v(t)), the rate of gasoline consumption as a function of time. determine using technology, the time when the car is running most economically during a 4-hour trip.

i conbined the functions and got c(v(t)) = ((40+3t+t^2)/500-0.1)^2+0.15 and i sub in the 4 hours in t.. and got 0.1507 what is that?
please help..
thanks in advance..

 

[wjm11]

the speed of a car, v kilometres per hour, at a time of t hours is represented by v(t) =40 +3t -t^2. the rate of gazoline consumtion of the car, c litres per kilometre, at a speed of v kilometres per hour is represented by c(v) =(v/500-0.1)^2 +0.15. i conbined the functions and got c(v(t)) = ((40+3t+t^2)/500-0.1)^2+0.15

You need a “-t^2” in your combined equation.

To find the lowest value of c(v(t)) between 0 and 4 hours, find the derivative and set it equal to zero.

Solve for t. Only use values for t between 0 and 4.

Plug that t value back into your original equation and solve for c.

Also plug 0 and 4 into the original equation and find those c values as well.

The smallest of the c values is the most economical rate.

Alternatively, graph the c(v(t)) function on your calculator and find the minimum value between 0 and 4 that way.