# Physics kinematics question (constant acceleration)

• TheDog
In summary, the police car begins emergency braking at 5 m/s2, which reduces the separation between the two cars to 14.6 m by the time your attention returns. If you had begun braking at 5 m/s2 too, you would have hit the police car at 96.1 km/h.
TheDog
1.You are arguing over a cell phone while trailing an unmarked police car by 29 m. Both your car and the police car are traveling at 105 km/h. Your argument diverts your attention from the police car for 2.0 s. At the beginning of that 2.0 s, the police officer begins emergency braking at 5 m/s2.
(a) What is the separation between the two cars when your attention finally returns?
(b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

2. v=v0+at
x-x0=v0t+(1/2)at2
v2=v02+2a(x-x0)
x-x0=1/2(v0+v)t
x-x0=vt-1/2at2

3. Part (a) gave me no trouble. 105 km/h=29.2 m/s

To find the position of the police car, I used equation 2 (x0=29 m) to find the position of the police car at t=2.0 and t=2.4 (the time at which you start braking). At t=2.0, I found the separation to be 19 m (correct). At t=2.4, I found the separation to be 14.6 m (used eqn. 2 to find x of the police car and your car and subtracted).

After 2.4s of braking, I found the velocity of the police car to be 17.2 m/s, so v(t)=17.2-5t (according to equation 1). Your velocity when you begin braking is v(t)=29.2-5t.

I tried using equation 3 to find v when you hit the police car: x-x0=14, substituting the values above I found it to be approximately 26.6 m/s or 96.1 km/h. However, this is not correct. I feel like it has something to do with a system of equations (to account for the different velocities) but I've been at this for close to two hours now and can't seem to wrap my head around it.

Any and all help is greatly appreciated.

Hi Dog, so Ill assume all you values are correct as you say they are for the first bits. I think the best way to approach this problem is be very consistent, from the looks of it your hoping to plug the values in a get an answer which isn't always easy to visualize, so let's try this way.

first let's call the position of the car at time t $x_c(t)$ and the position of the police car at time t $x_p(t)$. now we then can come up with an equation that describes the distance between these two cars:

$$s = x_p(t) - x_c(t)$$

ohk. Now we need to describe the positions of these two cars. so we know that the position of a body is equal to its initial position plus is displacement over a time t. so in equation form we have:

$$x_p(t) = (v_{p,0})t + \frac{1}{2}at^2 + x_{p,0}$$
$$x_c(t) = (v_{c,0})t + \frac{1}{2}at^2 + x_{c,0}$$

now you can see here that the equations above are simply a rearrangement of equation two you have listed below. Hopefully the use of the subscript letters and numbers isn't confusing, its to try and keep things tidy and clean but do say if it doesn't make sense and ill try to explain it better.

Putting this back together in the first equation we get:

$$s = [(v_{p,0})t + \frac{1}{2}at^2 + x_{p,0} ] - [(v_{c,0})t + \frac{1}{2}at^2 + x_{c,0} ]$$
$$= (v_{p,0})t + x_{p,0} - (v_{c,0})t - x_{c,0}$$
$$s = t (v_{p,0}- v_{c,0}) + x_{p,0}- x_{c,0}$$

right now hopefully you can see that this is a linear equation in t and s. now we actaully have all of the values for velocities and distances,

$$v_{p,0} = 17.2 m/s, \ v_{c,0}=29.2 m/s, \ x_{p,0} = 14.6 m, \ x_{c,0} = 0m$$

so plug those values in, do all you adding subtraction etc and we get:

$$s = 14.6 - 12t$$

now if you remember back to the beginning of my essay (:D) we said that s was the distance between the police car and normal car. But when they collide this distance will be 0, so let's set s to 0, and we get:

$$12t = 14.6$$

now try and go from here, what you'll get is a value of t, representing the time after the normal car began to brake, and you want the final velocity of the normal car, which means you will need to look back at considering this value of t with the deceleration of the normal car and its velocity when it began to brake. Hopefully that has help you and not confused you, good luck with finishing the question :D

Thanks a ton for the help. Using your equation I was able to get the right answer on my first try, and now I see where my error was, for some reason I just couldn't get there (the reason is probably that I'm a chemistry student and physics has always caused me endless headaches).

Thanks again for the help, I'm sure it won't be the last time I need it.

Hey Dog, ah that's great I am really glad that's helped you, I am a math/physics guys my self, and can totally see where you coming from, some of the stuff in chemistry i just push to one side :D

## 1. What is kinematics in physics?

Kinematics in physics is the study of motion and its causes, without taking into consideration the underlying forces responsible for the motion. It involves describing the position, velocity, and acceleration of an object over time.

## 2. What is constant acceleration?

Constant acceleration is when an object's velocity changes at a constant rate over time. This means that the object's acceleration remains the same throughout its motion.

## 3. How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 4. Can an object have constant acceleration if its speed changes?

Yes, an object can have constant acceleration even if its speed changes, as long as the change in speed is at a constant rate. This is because acceleration is not dependent on an object's speed, but rather on the change in its speed over time.

## 5. How does constant acceleration affect an object's motion?

Constant acceleration causes an object's velocity to change at a constant rate, which in turn affects its position and ultimately its motion. An object with constant acceleration will have a linear change in its position over time, resulting in a curved path known as a parabola.

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