Physics Homework Question: A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s2

[Amelinator]

A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.26 m/s2. How fast is the racer moving 334.0 m after the parachute opens?


My thought is that to find out how fast the car is going after the parachute opens, you need to find the final velocity before the parachute opens first. I'm just not sure how to do that.

 

[stapel]

A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.26 m/s2. How fast is the racer moving 334.0 m after the parachute opens?

View attachment 37032
My thought is that to find out how fast the car is going after the parachute opens, you need to find the final velocity before the parachute opens first. I'm just not sure how to do that.

What relationship does velocity have to acceleration?

 

[khansaheb]

A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.26 m/s2. How fast is the racer moving 334.0 m after the parachute opens?

View attachment 37032
My thought is that to find out how fast the car is going after the parachute opens, you need to find the final velocity before the parachute opens first. I'm just not sure how to do that.

There are 3 equations that relate initial velocity (u), final velocity (v), constant acceleration (a), distance travelled (s) and time of travel (t), of a particle that is travelling in a straight line. These are the three laws of linear particle motion under constant acceleration.

 

[Amelinator]

What relationship does velocity have to acceleration?

Acceleration is the rate of change of velocity. Right?

 

[khansaheb]

you need to find the final velocity before the parachute opens first

Correct...

You know the initial velocity(u = 0), you know the acceleration (a = 15.2 m/s²), you know the distance traveled (s = 401 m) - your (interim) "find" is final velocity (v).The pertinent equation should be:

v² = u² + 2*a*s

Continue.....

 

[The Highlander]

A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.26 m/s2. How fast is the racer moving 334.0 m after the parachute opens?

View attachment 37032
My thought is that to find out how fast the car is going after the parachute opens, you need to find the final velocity before the parachute opens first. I'm just not sure how to do that.

Hi @Amelinator,

Whenever I am teaching Physics (or indeed Maths, for similar types of problems) I exhort my pupils to always follow the same procedure.

In dealing with problems involving the Equations of Motion it's important, obviously, that the student knows these; we teach them (here, in Bonnie Scotland) as:-

$v=u+at\\s=ut+\frac{1}{2}at^2\\and\\v^2-u^2=2as$​

which @khansaheb has very kindly rearranged for you to: $\displaystyle v^2=u^2+2as$

I see you are following something along similar lines to what I teach but note that you are using v₀ for initial (or starting?) velocity instead of $\displaystyle u$ and "D" for displacement (or distance?) instead of $\displaystyle s$.
(I would discourage the use of the names I've put in brackets. )

You're free to use whatever you like, of course, but I would advise sticking to the variable names used in the original formulae; just to prevent any potential confusion.

I would also advise you to follow this procedure for every calculation you embark upon involving the Equations of Motion (EoM):-

0. Draw a sketch (if appropriate/required).​

1. List all the variables. (Include all five for EoM calcs. ie: "$\displaystyle u~v~a~s~t$")​

2. Assign values to those you know.​

3. Identify (Tick?) the unknown value you need to find.​

4. Choose (and write down) the formula that includes your known values and your desired unknown.​

5. Substitute your known values into the chosen equation.​

6. Rearrange (if reqd.) and solve the equation to get your desired value.​

So, for your initial part of the problem (the bit you say you're struggling with), I would write down:-

$u=0\\ v= ~?~\checkmark\\a=15.2~ms^{-2}\\s=401~m\\t=~?$​

Then, choosing, $\displaystyle v^2-u^2=2as$, I would write:-

$\displaystyle      v^2-u^2=2as\\ \implies v^2-0^2=2\times 15.2\times 401\\ \implies v^2=0+12,190.4\\ \implies v=\sqrt{12,190.4}\\ \implies \underline{\underline{v=~??~ms^{-1}}}$​

(Always double underline your final answers. )

And be sure to include the Units (like ms^-1) in any "final" answers!​

If you follow that procedure throughout you should have no further difficulties in reaching the final solution required by the problem as given.

(I would also call a negative value for $\displaystyle a$ "decelleration" but that's perhaps just semantics. )

Please now come back and show us your working as employed to reach the final solution requested by the problem as stated.
Thank you.

Hope that helps.

 

[Amelinator]

Correct...

You know the initial velocity(u = 0), you know the acceleration (a = 15.2 m/s²), you know the distance traveled (s = 401 m) - your (interim) "find" is final velocity (v).The pertinent equation should be:

v² = u² + 2*a*s

Continue.....

Sorry, my work is sloppy, but this is what I ended up getting. Thank
you for your help!

 

[Amelinator]

Hi @Amelinator,

Whenever I am teaching Physics (or indeed Maths, for similar types of problems) I exhort my pupils to always follow the same procedure.

In dealing with problems involving the Equations of Motion it's important, obviously, that the student knows these; we teach them (here, in Bonnie Scotland) as:-

$v=u+at\\s=ut+\frac{1}{2}at^2\\and\\v^2-u^2=2as$​

which @khansaheb has very kindly rearranged for you to: $\displaystyle v^2=u^2+2as$

I see you are following something along similar lines to what I teach but note that you are using v₀ for initial (or starting?) velocity instead of $\displaystyle u$ and "D" for displacement (or distance?) instead of $\displaystyle s$.
(I would discourage the use of the names I've put in brackets. )

You're free to use whatever you like, of course, but I would advise sticking to the variable names used in the original formulae; just to prevent any potential confusion.

I would also advise you to follow this procedure for every calculation you embark upon involving the Equations of Motion (EoM):-

0. Draw a sketch (if appropriate/required).​

1. List all the variables. (Include all five for EoM calcs. ie: "$\displaystyle u~v~a~s~t$")​

2. Assign values to those you know.​

3. Identify (Tick?) the unknown value you need to find.​

4. Choose (and write down) the formula that includes your known values and your desired unknown.​

5. Substitute your known values into the chosen equation.​

6. Rearrange (if reqd.) and solve the equation to get your desired value.​

So, for your initial part of the problem (the bit you say you're struggling with), I would write down:-

$u=0\\ v= ~?~\checkmark\\a=15.2~ms^{-2}\\s=401~m\\t=~?$​

Then, choosing, $\displaystyle v^2-u^2=2as$, I would write:-

$\displaystyle      v^2-u^2=2as\\ \implies v^2-0^2=2\times 15.2\times 401\\ \implies v^2=0+12,190.4\\ \implies v=\sqrt{12,190.4}\\ \implies \underline{\underline{v=~??~ms^{-1}}}$​

(Always double underline your final answers. )

And be sure to include the Units (like ms^-1) in any "final" answers!​

If you follow that procedure throughout you should have no further difficulties in reaching the final solution required by the problem as given.

(I would also call a negative value for $\displaystyle a$ "decelleration" but that's perhaps just semantics. )

Please now come back and show us your working as employed to reach the final solution requested by the problem as stated.
Thank you.

Hope that helps.

Thank you so much

 

[The Highlander]

View attachment 37042
Sorry, my work is sloppy, but this is what I ended up getting. Thank
you for your help!

Yes, it is a bit messy though I've seen worse!

However I am more concerned that you have shown the displacement as negative in your calculation!

As shown, the answer you got would not be the correct result to the calculation (even though it is the right answer to the problem)!

You need to be very careful with signs and you haven't (despite my BIG warning) included the units (ms^-1 or even m/s) in your final answer!

As a Physics teacher I would definitely deduct marks for that mistake!

​

 

[Amelinator]

Yes, it is a bit messy though I've seen worse!

However I am more concerned that you have shown the displacement as negative in your calculation!

As shown, the answer you got would not be the correct result to the calculation (even though it is the right answer to the problem)!

You need to be very careful with signs and you haven't (despite my BIG warning) included the units (ms^-1 or even m/s) in your final answer!

As a Physics teacher I would definitely deduct marks for that mistake!

View attachment 37047​

I will definitely be more careful next time! Thank you so much for all of your help.