How Do You Calculate When One Moving Object Overtakes Another?

[kittykatxox]

kinematics question help please :)?

So I've tried to attempt this problem by finding possible things like time or velocity but it just does not come out right. Help is greatly appreciated :)

Problem:

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.0 m/s^2. At the same instant a truck is 13.4 m down the road and traveling with a constant speed of 9.3 m/s. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

heres some work I did:
Truck:
v=x/t
9.3=13.4/t
t= 1.44 sThanks :)

 

[N-Gin]

If you know how far are the truck and the car from the traffic light at any moment, this problem is easy to solve.

So, for the car you have

$$x(t)=\frac{1}{2}at^2.$$

And for the truck

$$x(t)=x_0+vt.$$

Now you just need to figure what is $x_0$ and what is the condition for the car and the truck to meet.

 

[kittykatxox]

What is x(t)?

 

[N-Gin]

What is x(t)?

$x(t)$ is the function with $t$ as a variable. It tells you what is the distance $x$ from the traffic light at the time $t$.

 

[rwisz]

Hi there,

I can definitely help you with this kinematics question! Let's break it down and go through each part step by step.

First, we need to determine the time it takes for the truck to travel 13.4 m. You correctly used the equation v=x/t, where v is the velocity, x is the distance, and t is the time. Plugging in the given values, we get t=1.44 s.

Next, we need to find the distance the automobile travels in the same amount of time. We can use the equation x=1/2at^2, where a is the acceleration and t is the time. Plugging in the given values, we get x=1/2(2.0 m/s^2)(1.44 s)^2=2.61 m.

Now, we know that the automobile will overtake the truck at a distance of 13.4 m + 2.61 m = 15.01 m beyond the traffic signal.

For part (b), 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. Since the automobile starts from rest, v0=0. Plugging in the given values, we get v=0+2.0 m/s^2(1.44 s)=2.88 m/s. So at the instant the automobile overtakes the truck, it will be traveling at a speed of 2.88 m/s.

I hope this helps! Let me know if you have any further questions or need clarification on any of the steps. Good luck!