Car stopping time and distance when speed is doubled

[Just_enough]

Homework Statement

A. A car moving at speed v can stop in distance d in time t. The same car moving on the same road at twice the speed takes what distance and what time to stop?
Answer: a) Distance d and time t/2. b) Distance 2d and time t. c) Distance 4d and time 2t. d) Distance 2d and time 2t. e) Not possible to determine.

Homework Equations

A. V_f = V_i+2ad?
B. W = F*D?

The Attempt at a Solution

A.I assume that you would have to use velocity equation to figure this out, but I don't understand how you can use the equation if there is no numbers given...
B. I tried to find the work of the obj then set it equal to the work done on the person (pulling) and divide by force to find distance, and got 20... is this not the right equation?

 

[TomHart]

Problem A:

A. Vf = Vi+2ad?

It should be: v_f² = v_i² + 2ad
That is a good equation to use for distance.

For these kinds of problems, you just have to start by writing out the equations using the variables (no values) for the two different cases - case 1 and case 2 - using appropriate subscripts for case 1 and 2. You also have some additional information that the initial velocity in case 2 is twice that of case 1, which you can put into a simple equation and substitute into one of the other two equations. So what is the key variable that you are interested in? In this case it is d, so you want to solve both equations for d. Then in order to find the ratio of the two distances, you can divide one equation by the other. So you should end up with d2/d1 equal to some numeric value. That's how you have to solve these kinds of ratio problems. Then you will have to use a different equation to find the ratios of the times t₁ and t₂.

 

[Just_enough]

Problem A:

It should be: v_f² = v_i² + 2ad
That is a good equation to use for distance.

For these kinds of problems, you just have to start by writing out the equations using the variables (no values) for the two different cases - case 1 and case 2 - using appropriate subscripts for case 1 and 2. You also have some additional information that the initial velocity in case 2 is twice that of case 1, which you can put into a simple equation and substitute into one of the other two equations. So what is the key variable that you are interested in? In this case it is d, so you want to solve both equations for d. Then in order to find the ratio of the two distances, you can divide one equation by the other. So you should end up with d2/d1 equal to some numeric value. That's how you have to solve these kinds of ratio problems. Then you will have to use a different equation to find the ratios of the times t₁ and t₂.

so basically I plug in what info I have for each system, then solve for d and divide each equation by it's self?

 

[Any Help]

There is no need for numbers
use the simple equation d=vt
now duplicate v so that V=2v....V, D and T will be the new velocity, distance and time
now try to substitute in every choice V and the time given
a) D=V.T D=2v*t/2=v/t=d ..... so d and t/2 is correct
and continue in that way

 

[TomHart]

use the simple equation d=vt

d = vt applies for a constant velocity (acceleration = 0). This situation is a car that is decelerating - braking from some initial velocity to a stop (v=0).

so basically I plug in what info I have for each system, then solve for d and divide each equation by it's self?

I wouldn't say "divide each equation by itself"; I would say "divide one equation by the other".

 

[PeroK]

so basically I plug in what info I have for each system, then solve for d and divide each equation by it's self?

It is important to learn to work with algebra not just numbers. However, let me show you one thing that helps in these problems.

Just pick any speed for $v$ and any braking deceleration $a$ and calculate $d$ and $t$.

You might as well pick some easy numbers, so try $v = 20m/s$ and $a= -10m/s^2$.

Then repeat the calculation using twice the speed $v = 40m/s$ and calculate $d$ and $t$ for this case.

That will actually give you the answer - you just need to compare the answers in the two cases. Or, you could use this method to check an answer you got using algebra.

 

[Just_enough]

d = vt applies for a constant velocity (acceleration = 0). This situation is a car that is decelerating - braking from some initial velocity to a stop (v=0).I wouldn't say "divide each equation by itself"; I would say "divide one equation by the other".

how can you tell it decelerating?

 

[TomHart]

how can you tell it decelerating?

Because of the following wording in the problem statement:

A car moving at speed v can stop in distance d in time t.

The car goes from some initial velocity, v, to being stopped (v=0).

 

[Just_enough]

Because of the following wording in the problem statement:

The car goes from some initial velocity, v, to being stopped (v=0).

but you don't have to take that into consideration if it doesn't mention that do you?

 

[TomHart]

but you don't have to take that into consideration if it doesn't mention that do you?

I'm sorry, but I don't understand the point you are making in your statement above. If the problem doesn't mention that the car goes from some initial velocity to velocity = 0, then that would certainly have to be considered. But if that point was removed, it would be a different kind of problem altogether and more information would be needed.

This problem is an exercise in determining stopping distance relative to initial velocity. A car is moving along at some initial velocity and the driver slams on the brakes. How much time is required for it to stop, and what distance is required for it to stop. And how does that time and distance compare to that same car if its initial velocity is doubled? One might be inclined to think, "If the speed is doubled, the stopping distance would also be doubled and the time required to stop would be doubled." So if one works out the equations, that assumption will be proven or disproven.

 

[Just_enough]

I'm sorry, but I don't understand the point you are making in your statement above. If the problem doesn't mention that the car goes from some initial velocity to velocity = 0, then that would certainly have to be considered. But if that point was removed, it would be a different kind of problem altogether and more information would be needed.

This problem is an exercise in determining stopping distance relative to initial velocity. A car is moving along at some initial velocity and the driver slams on the brakes. How much time is required for it to stop, and what distance is required for it to stop. And how does that time and distance compare to that same car if its initial velocity is doubled? One might be inclined to think, "If the speed is doubled, the stopping distance would also be doubled and the time required to stop would be doubled." So if one works out the equations, that assumption will be proven or disproven.

what I mean is that for example, if a problem didnt mention about friction, then you don't need to calculate for friction too, but I see what you mean. also I tried to use the original equation posted and set the the 2x velocity to be 2v an device an equation by the other and I get either 1/2 or 2 since the variables cancels out...

 

[TomHart]

I get either 1/2 or 2 since the variables cancels out...

You get 1/2 or 2 for what parameter - distance or time?

 

[Just_enough]

You get 1/2 or 2 for what parameter - distance or time?

either one, unless I am doing it wrong.
v_f² = v_i²+2ad
----------------------------------------------------------- (divide)
v_f² = 2v_i²+2ad

the velocity, accel, and distance cancel out (doesnt really matter which variable I bring to the left side since It'll end up being the same). I think I am translating what they mean by diving the equation by the other wrong...

 

[TomHart]

v_f² = 2v_i²+2ad

It should be v_f² = (2v_i)² + 2ad

 

[Just_enough]

It should be v_f² = (2v_i)² + 2ad

Ok, I am still stump... I don't know what I am doing wrong. but I'm ending up with 1/4 (same as before where the var cancels out... unless I am doing my algebra wrong...

 

[TomHart]

I think 1/4 (or 4) is the right answer for distance. As the velocity doubles, the distance required to stop increases by a factor of 4.
So what about time? How much does it increase as velocity is doubled?

 

[Just_enough]

I think 1/4 (or 4) is the right answer for distance. As the velocity doubles, the distance required to stop increases by a factor of 4.
So what about time? How much does it increase as velocity is doubled?

using v_f=v_i+2at, I got 2 or 1/2 (idk which divide by which)

 

[TomHart]

Yes, 2 looks like the right answer for time. It's pretty easy to figure out which time is longer. Also, it's a good idea to use subscripts so you can keep track of things. So you might write:

v_1f = v_1i + 2at₁
v_2f = v_2i + 2at₂

Of course, the final velocity will be 0 in both cases.
And for case 2, the initial velocity is twice that of case 1, so v_2i = 2v_1i
Then when you divide that out you should get t₂/t₁ = 2

 

[Any Help]

d = vt applies for a constant velocity (acceleration = 0). This situation is a car that is decelerating - braking from some initial velocity to a s

Ah sorry I didn't pay attention for that