Distance/Time Equation - Seemingly Simple but possible trick question?

[Raptor]

Hello, I am stumped on a Math problem that seems to be a trick question. Here goes:

A student measures the distance in miles from home to school during driving as her Calculus project. Let s(t) be the distance from home at the instant t hours. She finds the function ?(?)=(((?−2)^2)/10)+2?, clearly, s(0) = 0 when she starts leaving from home at t = 0.
a. Find her average speed from time t = 12 hours to t = 15 hours
b. Find her instantaneous speed at time t = 10.
c. Based on this s(t), did she stop during her trip (at a stop sign perhaps)? When did she stop?
d. What is the distance from the stop sign to her home?

When I plug in the equation, I get this: https://www.desmos.com/calculator/sref9p1eog

I've derived the equation and got 0.2t +1.6

The first question I got 4.3 miles per hour just using the rate of change equation.

The second question I plugged in 10 for t to find the instantaneous rate of change at 3.6 miles per hour.

My problem involves the last two questions, c and d. The student starts at .4 miles away from home when time starts. The function carries on and he never stops. I'm not looking at quadrant 2 or 3 because that would mean I'm going 'back in time'.

I want to say that given the function the student never stops, however when talking to my teacher he said there is an 'algebraic trick' to finding out, and that I would be given partial credit if I say the student never stops.

Please take a look and let me know where I'm going wrong. Thanks for your help.

 

[mmm4444bot]

Hello, I am stumped on a Math problem that seems to be a trick question. Here goes:

A student measures the distance in miles from home to school during driving as her Calculus project. Let s(t) be the distance from home at the instant t hours. She finds the function ?(?)=(((?−2)^2)/10)+2?, clearly, s(0) = 0 when she starts leaving from home at t = 0.
a. Find her average speed from time t = 12 hours to t = 15 hours
b. Find her instantaneous speed at time t = 10.
c. Based on this s(t), did she stop during her trip (at a stop sign perhaps)? When did she stop?
d. What is the distance from the stop sign to her home?

When I plug in the equation, I get this: https://www.desmos.com/calculator/sref9p1eog

I've derived the equation and got 0.2t +1.6

The first question I got 4.3 miles per hour just using the rate of change equation.

The second question I plugged in 10 for t to find the instantaneous rate of change at 3.6 miles per hour.

My problem involves the last two questions, c and d. The student starts at .4 miles away from home when time starts. The function carries on and he never stops. I'm not looking at quadrant 2 or 3 because that would mean I'm going 'back in time'.

I want to say that given the function the student never stops, however when talking to my teacher he said there is an 'algebraic trick' to finding out, and that I would be given partial credit if I say the student never stops.

Please take a look and let me know where I'm going wrong. Thanks for your help.

The statement in red above is a typo, yes? You implied later that s(0)=0.4

I think the given function is a silly model for driving distance versus time.

Her average speed over 3 hours (from t=12 to t=15) is only 4.3 mph?!

The first derivative shows that her speed was zero only for an instant (exactly 8 hours before time zero), and then it takes eight hours for her speed to continuously increase from 0mph to 1.6mph! That's goofy.

Also, as posted, we cannot tell what t=0 means in context. Are you sure that you copied the problem correctly?

 

[sean]

Hello, I am stumped on a Math problem that seems to be a trick question. Here goes:

A student measures the distance in miles from home to school during driving as her Calculus project. Let s(t) be the distance from home at the instant t hours. She finds the function ?(?)=(((?−2)^2)/10)+2?, clearly, s(0) = 0 when she starts leaving from home at t = 0.
a. Find her average speed from time t = 12 hours to t = 15 hours
b. Find her instantaneous speed at time t = 10.
c. Based on this s(t), did she stop during her trip (at a stop sign perhaps)? When did she stop?
d. What is the distance from the stop sign to her home?

When I plug in the equation, I get this: https://www.desmos.com/calculator/sref9p1eog

I've derived the equation and got 0.2t +1.6

The first question I got 4.3 miles per hour just using the rate of change equation.

The second question I plugged in 10 for t to find the instantaneous rate of change at 3.6 miles per hour.

My problem involves the last two questions, c and d. The student starts at .4 miles away from home when time starts. The function carries on and he never stops. I'm not looking at quadrant 2 or 3 because that would mean I'm going 'back in time'.

I want to say that given the function the student never stops, however when talking to my teacher he said there is an 'algebraic trick' to finding out, and that I would be given partial credit if I say the student never stops.

Please take a look and let me know where I'm going wrong. Thanks for your help.

Ok first off @ t = 0:

s(0) = (((0-2)^2)/10) + 2(0)

Which leads to the distance at t = 0 (s(0)) being:

s(0) = 4/10 or 0.4 (this means that for some reason she is already .4miles closer to home before even setting out)

(a) Next to find her average speed. You get distance at s(12) and s(15) and take them away from each other. This distance was traveled in 3 seconds (15 - 12) and then you just use the simple formula:

distance/time = speed

(b) Instance speed at t = 10 simply means you fill in 10 for t as follows:

s(10) = (((10-2)^2)/10) + 2(10)

and solve

(c) This really means at any point is her speed 0 miles/hour (did she stop)

For which the answer is yes. We know that speed = distance/ time

So we can say speed at a certain time t (v(t)) is equal to the following

v(t) = ((((t-2)^2)/10) + 2(t))/t

Working this out you get

v(t) = ((t-2)^2)/10t + 2

Then let v(t) = 0 and work value for t out. This t is when she stopped.

(d) Then plug that value of t into s(t) formula and then take that distance from s(15) and that is how far away it is from the house.

 

[mmm4444bot]

Ok first off @ t = 0:

s(0) = (((0-2)^2)/10) + 2(0)

Which leads to the distance at t = 0 (s(0)) being:

s(0) = 4/10 or 0.4 (this means that for some reason she is already .4miles closer to home before even setting out)

Raptor already did this calculation. See the original post.

When you say, "closer to home", what does that mean? Closer than what? Function s(t) gives the distance FROM home (in miles).

(a) Next to find her average speed. You get distance at s(12) and s(15) and take them away from each other. This distance was traveled in 3 seconds (15 - 12) and then you just use the simple formula:

distance/time = speed

Raptor did this calculation already. The original post shows the result: 4.3mph

The units for t are given as hours, not seconds.

(b) Instance (sic) speed at t = 10 simply means you fill in 10 for t as follows:

s(10) = (((10-2)^2)/10) + 2(10)

and solve

This is not correct, Sean. The original post asks for the "instantaneous" speed at t=10 hours. Function s(t) does not output a speed; it outputs a distance.

Your work above results in the distance from the student's house ten hours after time zero.

(c) This really means at any point is her speed 0 miles/hour (did she stop)

For which the answer is yes. We know that speed = distance/ time

So we can say speed at a certain time t (v(t)) is equal to the following

v(t) = ((((t-2)^2)/10) + 2(t))/t This is not correct.

s(t) is a displacement function.

v(t) is the first derivative of displacement.

Working this out you get

v(t) = ((t-2)^2)/10t + 2

Then let v(t) = 0 and work value for t out. This t is when she stopped.

Raptor already calculated the first derivative. See the original post.

v(t) = 0.2t + 1.6

This velocity function is zero at only one value (t=-8).

(d) Then plug that value of t into s(t) formula and then take that distance from s(15) and that is how far away it is from the house.

For part (d), why are you using the distance from the house at t=15?

I think that we need to wait for Raptor to confirm the given displacement function. What's posted does not make sense, in the context of the given scenario (eg: 8 hours to accelerate from 0 to 1.6 miles per hour). :cool: