Decelerating Car Anti-Derivatives Question

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

A car is travelling at 80 km/hr when the brakes are fully applied producing a constant deceleration of 7 m/s^2. Use antiderivatives to find the distance travelled before the car comes to a stop.

Based on what the question is requesting, a(t), v(t), and d(t) must be computed in order to compute the total distance travelled. Although one can easily solve this via a Physics formula, the anti-derivative method is suggested.

Therefore, I converted v(0) = 80km/hr to 22.222 m/s (repeating)

Also, due to deceleration, acceleration is expressed as the following function:

a(t) = -7

Taking the anti-derivative gives:

v(t) = -7t + 22.222 and found d(t)

d(t) = (-7/2)t^2 + 22.222 + D.

At this point, I didn't know what to do. I used the following formula, but I'm somewhat doubtful as to whether this is "allowed" since strictly anti-derivatives are requested:

v=vo + at

Therefore,
t=3.174 seconds.

D= V*t

Thus,
d= 22.222 + 3.174
d= 35.3 metres travelled.

Am I close?

Thanks in advance.

 

[Ishuda]

You are correct in what you were doing. You just sort of changed notations in the middle. Lets start at the point where you have
v(t) = -7t + 22.222
so your vo = 22.222 and a = -7.

Taking the anti-derivative gives
d(t) = -3.5 t^2 + 22.222 t + D.
That is, 22.222 is multiplied by t. I think this might be where you made a typo (or possibly a mistake) and got confused. At time t=0, d(t) is zero, so D is zero and, with a little rearrangement,
d(t) = ( -3.5 t + 22.222 ) t

I get a slightly different answer for when v is zero, t is 3.1746 (a matter of rounding differences), so that d(3.175) = 35.273

 

[HallsofIvy]

It might be better to use $\displaystyle \frac{200}{9}$ rather than 22.222... Also, since you are asked for "distance traveled while decelerating", you can take the initial position to be 0- the constant, D, in $\displaystyle d(t)= -\frac{7}{2}t^2+ \frac{200}{9}t+ D$ is 0. Solve $\displaystyle v(t)= -7t+ \frac{200}{9}= 0$ to determine when the car has come to a stop and put that into the formula for d(t) to find the distance.

The formulas you are not sure are "allowed" are precisely what you get by taking the anti-derivative!

 

[ardentmed]

It might be better to use $\displaystyle \frac{200}{9}$ rather than 22.222... Also, since you are asked for "distance traveled while decelerating", you can take the initial position to be 0- the constant, D, in $\displaystyle d(t)= -\frac{7}{2}t^2+ \frac{200}{9}t+ D$ is 0. Solve $\displaystyle v(t)= -7t+ \frac{200}{9}= 0$ to determine when the car has come to a stop and put that into the formula for d(t) to find the distance.

The formulas you are not sure are "allowed" are precisely what you get by taking the anti-derivative!

Thanks for the clarification. Your answer was very helpful.