Finding Acceleration Homework Solution

[mistabry]

Homework Statement

You are driving along a highway at 80.0 km/hr in car that is 6.0 m long. Exactly 10 m ahead of you is a truck of length 12.0 m driving at 60.0 km/hr.

a. If you wish to pass the truck so that the back of your car just clears the front of the truck in 3.0 s, how fast must you accelerate?
b. What will your speed be when you pass the front end of the truck?

Homework Equations

I thought I could use X_f-X_i = V_ixt + 1/2a(t)^2, but it didn't give me the correct answer.

The Attempt at a Solution

Plugged in X_f=28km V_ix=80km/h and t=3, but I couldn't get the correct answer. I didn't attempt B because I couldn't find A >.<

 

[housemartin]

If you are associating coordinate X with moving truck (truck's beginning or end), then Vi should be cars speed relative to the truck, not 80 [km/h]. Distances are given in meters, time in seconds, speed in kilometers per hour. So first make all units to be the same. Don't put different types of units into one equation.

 

[Aviation]

The best way would be to define your relative speed first which is 20 kmh. So you are already catching up with him. The distance to travel would be 28.0 m just like you said. 20 kmh = 5.55 m/s. Because you are already driving at a certain speed and you are accelerating you have to combine some factors. Theoraticly the surface between the V(t) and the x-axis is equal to the distance you travel. v(t) = at + 5.55
The intergral of this function describing your surface is equal to: s(t) = 1/2at² + 20t

if you fill in the variables: 28 = 0.5 * a * 9 + 3 * 5.55

And I think you can do the rest yourself.
In the end you'll have to use the following function: s = 1/2 at² + vt

 

[mistabry]

Thank you Aviation. I got the answers right with your method with a. 2.522 m/s^2 and b. 107.24 km/hr

 

[rwisz]

Hello! I would like to provide a response to your question regarding finding acceleration in this scenario.

Firstly, it is important to note that the equation you used, Xf-Xi = Vixt + 1/2a(t)^2, is a valid equation for finding the displacement of an object given its initial position, initial velocity, acceleration, and time. However, in this problem, we are not given the acceleration and we are trying to solve for it.

In order to find the acceleration, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this problem, we are given the initial and final velocities, as well as the time. Therefore, we can rearrange the equation to solve for acceleration, which gives us a = (v-u)/t.

Now, let's apply this equation to part a of the problem. We know that the initial velocity (u) is 80 km/h, the final velocity (v) is 60 km/h, and the time (t) is 3 seconds. Plugging in these values, we get a = (60 km/h - 80 km/h)/3 s = -6.67 km/h^2. Since acceleration is a vector quantity, the negative sign indicates that the car is decelerating.

For part b, we can use the equation v = u + at again. This time, we are solving for the final velocity (v). We know that the initial velocity (u) is 80 km/h, the acceleration (a) is -6.67 km/h^2, and the time (t) is 3 seconds. Plugging in these values, we get v = 80 km/h + (-6.67 km/h^2)(3 s) = 60 km/h. This means that the car will be travelling at 60 km/h when it passes the front end of the truck.

I hope this explanation helps you understand how to find acceleration in this scenario. Remember to always carefully identify the given variables and use the correct equations to solve for the unknown quantities. Keep up the good work in your studies!