Can't figure out this problem regarding two bodies with different accelerations

[NutriGrainKiller]

I am so close to figuring this problem out, but I just can't quite get it. Here's the problem:

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s^2. At the same instant a truck, traveling with a constant speed of 23.5 m/s, overtakes and passes the car.
A) How far beyond its starting point does the car overtake the truck?
B) How fast is the car traveling when it overtakes the truck?

Here is what is given:
car: Vyi = 0, Ax = 2.8 m/s^2
truck: Vyi = 23.5 m/s, Vyf = 23.5 m/s, Ax = 0

here is what i know:
1)If graphed, y-axis being distance and x-acis being time, the truck would like like a straight diagonal line, while the car would be sloping up, crossing over the truck's path at some point (I'll call it X). I am trying to find X.
2)I know I need to use two different kinematic equations and set them equal to each other, but this is where I start to not understand.
3) So, what I am trying to find is when the distance of both equations are equal to each other? So solve a couple kinematic equations for D then set them to each other right? Here's what I get:

(23.5 m/s)(T) = (1/2)(2.8 m/s^2)(T^2)

which turns out to be T = 16.79 seconds..is this correct? If it is I can use it to get the final answers for both A and B. Thanks guys!


please don't waste your time with this, turns out I was right!

 

[anathema]

Hello,

Thank you for sharing your thought process and progress so far. It looks like you are on the right track!

To find the distance (A) that the car overtakes the truck, you can use the equation d = v0t + 1/2at^2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity for the car is 0 and the acceleration is 2.8 m/s^2. The initial velocity for the truck is 23.5 m/s and the acceleration is 0. So, the equation for the car becomes d = 0 + 1/2(2.8)(t^2) and the equation for the truck is d = 23.5t.

Since we want to find the distance when the car overtakes the truck, we can set these two equations equal to each other and solve for t. This gives us:

0 + 1/2(2.8)(t^2) = 23.5t

Solving for t, we get t = 16.79 seconds, which confirms your answer.

To find the speed (B) of the car when it overtakes the truck, we can use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity for the car is 0 and the acceleration is 2.8 m/s^2. The initial velocity for the truck is 23.5 m/s and the acceleration is 0. So, the equation for the car becomes v = 0 + (2.8)(16.79) = 46.99 m/s.

Therefore, the final answers for A and B are:
A) The car overtakes the truck at a distance of 23.5(16.79) = 394.465 meters.
B) The car is traveling at a speed of 46.99 m/s when it overtakes the truck.

I hope this helps. Keep up the good work!

 

[quicknote]

Hi there,

It looks like you have a good understanding of the problem and how to approach it. You are correct in using kinematic equations to solve for the distance and time when the car overtakes the truck. Your equation for time, T = 16.79 seconds, is correct.

To solve for the distance, you can plug this time value into either of the kinematic equations you used and solve for distance. The answer should be around 197.6 meters.

To find the speed of the car when it overtakes the truck, you can use the kinematic equation Vf = Vi + at, where Vi is the initial velocity (which is 0 in this case), a is the acceleration of the car (2.8 m/s^2), and t is the time you found (16.79 seconds). This should give you a final velocity of 47.0 m/s.

I hope this helps! Keep up the good work and keep practicing with kinematic equations. They can be tricky at first, but with practice, you'll become more comfortable using them to solve problems like this one. Good luck!