Constant acceleration of sports car

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Homework Help Overview

The discussion revolves around a scenario involving constant acceleration in the context of a sports car braking to avoid hitting a dog. The original poster presents a problem where they need to determine the stopping distance and whether they can avoid a collision, given an initial velocity and a specified deceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss kinematic equations relevant to constant acceleration and the importance of unit consistency. There are attempts to convert units from kilometers per hour to meters per second and to apply the equations to find stopping distance and time.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning previous results. Some have provided guidance on unit conversions and the correct application of kinematic equations. There is recognition of earlier mistakes, and a few participants have arrived at consistent answers for stopping distance and time.

Contextual Notes

There are indications of confusion regarding the application of equations and unit conversions, which some participants are attempting to clarify. The original poster expresses frustration with their understanding of the topic, suggesting that the learning process is ongoing.

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Homework Statement




You are driving your new sports car at a velocity of 90km/hr , when you suddenly see a dog step into the road 50m ahead . You hit the breaks hard to get maximum deceleration of 7.5m/s² .



Homework Equations



i)How far will you go before stopping?

ii)Can you avoid hitting the dog?

can anyone help me ? please ?
 
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What equations do you know that involve kinematics?
 
the topic was constant acceleration .. then i hardly to understand .. it means i can't understand my teacher .. not only me .. >.<
 
Well the kinematic equations can be found on this webpage. Then read the paragraph that follows the equations, and try to understand them well because you'll need to use them a lot. These equations ONLY apply when acceleration is constant.

For this question you are told the:
Initial velocity, vi = 90 km/h
Final velocity, vf = 0 km/h (because you've stopped)
Acceleration, a = -7.5 m/s^2

So look for an equation that has all three of these values in order to get something useful :). Be careful with the units, and make sure you are consistent with length and time (i.e. make sure you always use meters and seconds, or kilometers and hours).
 
my answer was -150m .. and the true answer was 42m only ..

I think my brain will be explode ..

can you show how to solve EmittingLight ?

please ?
 
It's better if you show your working so that we can show where you went wrong.

I suspect it might be unit conversions, which you can check with tools like Wolfram Alpha. The initial velocity is 90 km/h whereas the acceleration uses meters and seconds so you have to convert one of them to the other's units. I like using SI units, so I'd convert to meters and seconds. In Wolfram Alpha you would type something like this. With all the unit conversions you should get:

Initial velocity, vi = 25 m/s
Final velocity, vf = 0 m/s (because you've stopped)
Acceleration, a = -7.5 m/s^2

If you look at the equations there are only two that have all of these variables:
v_{f}^2=v_{i}^2+2*a*dv_{f}=v_{i}+a*tYou know how to solve for t and d in the equations right? t stands for the time elapsed and d stands for the distance traveled. Do you know which equation you want then?
 
to solve for t:

i used the second eq.

for d:

i used the first eq .

my t is -17.5s

then to my d is -610m

then I divide the both to have velocity ..

what's next ? :?
 
Three of the values are known, I've told you what they are in all my previous posts: v_f, v_i and a.

I'll put these into the equations for you:0^2=25^2+2*-7.5*d0=25+-7.5*tDo you know how to solve for t and d now? Remember, they are all in meters and seconds now, since that's what I've converted them into.

Would you be able to show your working? Like show how you solved for t and d, so I can tell you what you did incorrectly.

EDIT: Nevermind I know what you're doing. You're not doing the algebra correctly. Do you know that if:
a=b*x
That:
\frac{a}{b}=x?

So if:
-625=2*-7.5*d
Then:
\frac{-625}{2*-7.5}=d
 
Last edited:
0^2=25^2+2∗−7.5∗d
0=625-15d
(transfer:)
15d=625
divide both sides by 15 to remain the d
then, d=41.67m

for the second:
0=25+-7.5*t
0=25-7.5t
(transfer:)
7.5t=25
divide both sides by 7.5 to remain t:
t=10/3s

is it wrong ?
 
  • #10
Well, didn't you say that the true answer was 42m before? It looks like it was rounded up from 41.67m to me.

Do you know why you got t=-17.5 and d=-610 before? And not now? Based on what I've deduced, you actually already put in the correct values before.

EDIT:
I just thought I should add that since you only wanted d, you didn't need to use the second equation. That's why I asked before if you 'knew which equation to use'; you wanted to find d, so you needed to use the first equation. But look on the bright side of this extra confusion, now you know how long it took the car to stop! And yes, you have the correct answers now as far as I know.
 
Last edited:
  • #11
I think I'm going to crazy .. but so very thank you .. now i know how the constant acceleration ..
 

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