Calculating Distance Traveled Using Antiderivatives: Solving for k

[yangxu]

Homework Statement

A car going 70km/h comes to a stop in 6 seconds, assume that the deceleration is constant, find the distance traveled using a graph; find the distance traveled using antiderivatives.

The Attempt at a Solution

If the deceleration is constant, I would assume that the slope (derivative) of the graph would also be constant, hence the graph should look like a straight line with a negative slope.

Converting 70km/h into m/s, I get 19.4m/s. To calculate the distance, I simply took the area under the curve, which is a triangle:

(19.4 x 6)/2 = 58.2m

But when I'm using antiderivatives to solve this, I get a different answer...

S(6) = intergral v(t)dt = 19.4 (6) = 116.4m

It seems like this answer is twice as much as the one when calculated using a graph, and I'm sure the graph should be the right answer... could someone please tell me which part did I do wrong?

 

[NateTG]

The velocity is not constant, but rather a function of time..

$$\int kt dt = \frac{1}{2} k t^2+C$$

 

[yangxu]

Thanks for your correction Nate. I'm still a bit confused though, if velocity at time 0 = 19.4, and that at time 6 is 0, wouldn't C always equal to 0?

 

[HallsofIvy]

Not "always". For this particular problem, yes.

 

[yangxu]

Thanks HallsofIvy, I attempted to solve using the corrected equation, but I still can't seem to get the right answer. I believe k is a constant in this formula, but how do I determine it? At time 0, the whole 1/2(kt^2) is equal to 0, am I supposed to use the distance at a different time (eg. t = 1s) to solve for k?