# How far did Andrew travel before stopping and did he hit the fox?

• -Dragoon-
In summary: I'm confused, now. d = \frac{V_i + V_f}{2}~\times~twhere [itex]V_i[/tex] is the initial velocity (20.8 m/s), [itex]V_f[/tex] is the final velocity (0 m/s), and t is the time (3.4 seconds).So, the distance he traveled before stopping is:d = \frac{20.8 + 0}{2}~\times~3.4 = \frac{20.8}{2}~\times~3.4 = 10.4~\times~3.4 = 35.36 \text{ m}Therefore, Andrew
-Dragoon-

## Homework Statement

Andrew is driving his van to deliver groceries in Elmvale. As he travels along the highway, a fox stops on the road. Andrew is traveling at 75km/h and is 42m away from the fox when he applies his brakes. It takes him 3.4 seconds to stop.
a) How far did Andrew travel before stopping?
b) Did Andrew hit the fox? Explain your answer by including a calculation.
c) If a typical van is 4.2m in length, how many van lengths did it take Andrew to stop?

## Homework Equations

Distance = V1 (delta t) + 1/2 (a)(delta t^2)
Distance = V2 (delta t) - 1/2 (a)(delta t^2)

## The Attempt at a Solution

For part a:
So I first convert 75km/h into m/s, to get 20.8 m/s. I know distance = velocity times change in time. Therefore, delta D = 20.8m/s(3.4 seconds) = 70.8m, is this really the distance he traveled before stopping?

For part b: You see, this is why I don't think my answer for part a) is correct. If it were, it is obvious he hit the fox because he travels 70.8m before hitting the fox and he was only 42m away when he applied the breaks. Why would they ask me this question if he didn't?

For part c: By using the answer I got in part a, it would be 70.8/4.2 correct? Which gives 16.8m, which is about 17 vans?

I don't believe I am correct, as I just learned a couple more complex equations but don't exactly know how to apply them. Thanks for all the help!

The distance you've calculated (for part a) is how far Andrew would have driven in 3.4 seconds if he decided he didn't care about the fox and didn't step on the brakes.

At the moment he steps on the brakes, he is going 75 km/hr, but 3.4 seconds later, he is going 0.0 km/hr (he has stopped). If you assume he decelerates at a constant rate, what is the acceleration that takes you from 75 km/hr to 0 in 3.4 seconds (what equation do you use)?

One way to get to the answer for part a from there is to use this acceleration and one of the equations you've already written that allows you to find distance traveled when you know the amount of time, the initial velocity, and the (constant) acceleration.

I hope that helps.

Retribution said:
So I first convert 75km/h into m/s, to get 20.8 m/s. I know distance = velocity times change in time. Therefore, delta D = 20.8m/s(3.4 seconds) = 70.8m, is this really the distance he traveled before stopping?
No it's not. You have calculated the velocity ok, but you didn't take acceleration into consideration. The car is traveling with velocity of 20.8 m/s and starts braking. After 3.4 seconds, velocity is zero. So therefore, you need v=v0 + a*t, where v is 0 (car stops) and v0 is 20.8. After that, you get acceleration of -6.314. It's negative while car is stopping. To get the distance you have to apply the next formula d=v0*t + (a*t^2)/2 (don't forget that acceleration is negative)

jamesrc said:
The distance you've calculated (for part a) is how far Andrew would have driven in 3.4 seconds if he decided he didn't care about the fox and didn't step on the brakes.

At the moment he steps on the brakes, he is going 75 km/hr, but 3.4 seconds later, he is going 0.0 km/hr (he has stopped). If you assume he decelerates at a constant rate, what is the acceleration that takes you from 75 km/hr to 0 in 3.4 seconds (what equation do you use)?

One way to get to the answer for part a from there is to use this acceleration and one of the equations you've already written that allows you to find distance traveled when you know the amount of time, the initial velocity, and the (constant) acceleration.

I hope that helps.

Oh, so first I must find acceleration? Ah, I forgot.

The formula I'll be using: V2 = V1 + a (delta t)
V2 = 0
V1 = 20.8m/s
change in time = 3.4 seconds
therefore, acceleration is: -6.12?

So, now I'll use the formula for distance: Distance = V1 (delta t) + 1/2 (a)(delta t^2)
Distance = 20.8m/s (3.4) +1/2 (-6.12m/s^2)(3.4^2)
Distance = 70.72 - 35.3736
Distance = 35.35m, is this correct?

method_man said:
No it's not. You have calculated the velocity ok, but you didn't take acceleration into consideration. The car is traveling with velocity of 20.8 m/s and starts braking. After 3.4 seconds, velocity is zero. So therefore, you need v=v0 + a*t, where v is 0 (car stops) and v0 is 20.8. After that, you get acceleration of -6.314. It's negative while car is stopping. To get the distance you have to apply the next formula d=v0*t + (a*t^2)/2 (don't forget that acceleration is negative)

How did you get an acceleration of -6.314? This just threw me off when I thought added, xD.

-20.8/3.4 = -6.12, no?

Retribution said:
How did you get an acceleration of -6.314? This just threw me off when I thought added, xD.

-20.8/3.4 = -6.12, no?

Ups...Well, I divided 75000/3600=20,8333 m/s and then i divided that with 3,4 and result is 6,1274, i.e. 6.13. I wrote wrong...

method_man said:
Ups...Well, I divided 75000/3600=20,8333 m/s and then i divided that with 3,4 and result is 6,1274, i.e. 6.13. I wrote wrong...

Ah, Ok. So, is my the distance 35.35m before he stopped?

Also, how would I answer b part by including a calculation? If the above is correct, all I would need to calculate is the distance he was away from the fox subtracted by the distance he traveled? 42m-35.35m?

Retribution said:
Ah, Ok. So, is my the distance 35.35m before he stopped?

Also, how would I answer b part by including a calculation? If the above is correct, all I would need to calculate is the distance he was away from the fox subtracted by the distance he traveled? 42m-35.35m?
Yes, that's how I would answer that.

There is no need to calculate the acceleration first. Just use the basic kinematic equation

$$d = \frac{V_i + V_f}{2}~\times~t$$

where [itex]V_i[/tex] is the initial velocity (20.8 m/s), [itex]V_f[/tex] is the final velocity (0 m/s) and t is the time (3.4 s)

## What is a simple kinematics problem?

A simple kinematics problem is a type of physics problem that involves the study of motion without considering the causes of that motion. It involves analyzing the position, velocity, and acceleration of an object using equations such as the equations of motion.

## What are the key concepts in solving a simple kinematics problem?

The key concepts in solving a simple kinematics problem are displacement, velocity, and acceleration. Displacement is the change in position of an object, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

## How do you approach solving a simple kinematics problem?

To solve a simple kinematics problem, you need to first identify the known and unknown variables, such as initial and final position, time, and acceleration. Then, use the appropriate kinematics equations to solve for the unknown variable. It is important to pay attention to units and use the correct equations for the given scenario.

## What are some common mistakes when solving a simple kinematics problem?

One common mistake is misunderstanding the meaning of the variables and using the wrong equations. Another mistake is not paying attention to units and using incorrect conversions. It is also important to be careful with algebraic manipulations and to double check calculations.

## How can simple kinematics problems be applied in real-life situations?

Simple kinematics problems can be applied in real-life situations, such as calculating the speed and distance of a car in a race, determining the acceleration of an object in free fall, or analyzing the motion of a projectile. They can also be used in fields such as sports, engineering, and transportation.

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