Question

A train travels along a track and its speed (in miles per hour) is given by s(t)=29t for the first half hour of travel. Its speed is constant and equal to s(t)=29/2 after the first half hour. Here time t is measured in hours. How far (in miles) does the train travel in the first hour of travel?

I keep getting 14.5 and I have tried multiple ways.

Answer 1

What class is this for? Calculus?

Answer 2

Pre-calculus

Answer 3

Hmm... That makes things a bit more difficult.
What section are you on?

Answer 4

our chapter title is called, Integrals, antiderivatives, and the Fundamental Theorem of CalculusIntegrals, antiderivatives, and the Fundamental Theorem of Calculus

Answer 5

Ahh, antiderivatives.
When working with position, speed, and acceleration, we know that position is the antiderivative of speed.

What's the antiderivative of the speed function?

Answer 6

I am looking at my notes and I do not quite understand the anti derivative so my guess would be 14.5t \[\int\limits_{0}^{1/2}\]?

Answer 7

The really simple rule with antiderivatives is:
\[ax^n \rightarrow \frac{a}{n+1}x^{n+1}\]

Answer 8

So:
\[29t^1 \rightarrow \frac{ 29 }{ 2 }t^2 = 14.5t^2\]

Answer 9

Usually there would be a +C, but we're assuming it starts at a position of zero.

Answer 10

so what would you do next?

Answer 11

So, we now have a position function that uses time as the input. How far would the train travel in the first 30 minutes?

Answer 12

so it would be 1/2 right?

Answer 13

Yeah, t=1/2

Answer 14

So then I would solve it right?

Answer 15

Ah finally got the answer thanks!