Grade 12 Kinematics -- Will the car stop before the cliff?

[lion1014]

Homework Statement

A car is traveling at 25.8 m/s when the driver notices that
the bridge is out 150 m ahead. If the driver applies the brakes
quickly with a uniform acceleration rate of - 2.55 m/s^2
, will the car stop before the cliff and if so how far from the edge will
the car be?
b. How long will it take for the car to stop through the last 35.6 m of
it's motion?

Relevant Equations

d = vi * t + 1/2at^2

t= v/a
=25.8 m/s / -2.5 m/s^2
=10.1 s

therefore,
d= vi * t + 1/2at^2
= (25.8)(10.1) +1/2(-2.55)(10.1)^2
=131 m
I got the first one right I think, but I don't know how to do b

 

[Delta2]

Hi and welcome to Physics Forums

I also think your answer for a) is correct (approach and equations used look correct), I was kind of lazy though (hehe) and didn't check the full arithmetic operations.

Now for b) do you know the equation $$V_f^2-V_i^2=2aD$$ where $V_f$ is the velocity at the end of the distance $D$ that is covered, $V_i$ is the velocity at the start and $a$ is the acceleration (or deceleration). If yes you know this equation, how it can be used to solve the problem?
If no, you got to tell me what other kinematic equations you know.

 

[gneill]

Hmm. @Delta2, I think that that SUVAT equation would be ideal for solving part (a), but a bit more involved for part (b). Since part (a) has already been solved by the OP I can offer this version for consideration:

Given $V_f^2 - V_i^2 = 2 a D$ and you want to find the stopping distance D, then since $V_f = 0$ ,
$$D = -\frac{V_o^2}{2a}$$
So, done and dusted with a single SUVAT equation.

For part (b) I would consider "playing the scenario in reverse", that is, suppose the car starts with zero velocity at its stopping point and accelerates backwards from there with the same magnitude of acceleration as before. That looks like a good case for @lion1014's relevant equation.

 

[kuruman]

I got the first one right I think, but I don't know how to do b

How far from the initial position of the car is the start of the beginning of the remaining 35.6 m ?
What is the velocity of the car at that point?

If you can answer these two questions, you should be able to get the answer using the same equations that you used for the first part.

 

[Delta2]

Yes well @gneill if I understood your approach well, we don't have to find the initial velocity as an intermediate step. However we would have to argue a bit additionally on why the time reversal scenario is equivalent. I don't think time reversal is within the scope of 12th grade kinematics.

 

[gneill]

Perhaps a thought experiment example? You drop (initial velocity zero) an object from some height H and it hits the ground after time t and with speed V. With what speed should you launch a projectile from the landing point to reach the same height H in time t and with zero remaining velocity? Clearly you'd launch it with speed V. It's just a time-reversed scenario.

 

[Delta2]

Perhaps a thought experiment example? You drop (initial velocity zero) an object from some height H and it hits the ground after time t and with speed V. With what speed should you launch a projectile from the landing point to reach the same height H in time t and with zero remaining velocity? Clearly you'd launch it with speed V. It's just a time-reversed scenario.

Ok well, I am just saying this whole idea of time reversed might not be so straightforward for a 12th grade student but then again I am not sure. What age group is the 12 grade in US again?

 

[gneill]

What age group is the 12 grade in US again?

I'd say 17 or 18 years old.

 

[Delta2]

I'd say 17 or 18 years old.

Ok , I thought it was for younger something like 13-15 years old. Well ok I guess its up to him/her to decide which method he/she finds easier.

 

[lion1014]

t= square root -2(35.6m)/ (-2.55) and I goy 5.28 s is this right?

 

[Delta2]

Yes it is correct.

 

[Delta2]

But how did u do it, using @gneill approach or using my approach (that is finding the velocity $V_i$ first)?

 

[lion1014]

Yes it is correct.

Thank you!