Solving the Traffic Light Overtake Puzzle

[tumbler19]

I have been working on this problem for a while, and I still don't understand how to set it up for the first part of this problem:

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s squared. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile.

(A) How far beyond the traffic signal will the automile overtake the truck?

(B) How fast will the automobile be traveling at that instant?

 

[Leong]

Use this formula,
$$s=ut+\frac{1}{2}at^2$$
Initial velocity for the car is u=0.
Constanst acceleration for the truck is a=0.
$$s_{car}=1.1t^2$$
$$s_{truck}=9.5t$$
s is the position relative to the origin / traffic light.

Part (a)
When the automobile overtakes the truck, their positions relative to the traffic light is the same. so, we have
$$s_{car}=s_{truck}$$. find the t when this happens and substitute back either of the equation for s above because they will be at the same position relative to traffic light.

Part (b)
Use this formula,
$$v=u+at$$

 

[M98Ranger]

First, it's important to note that this problem can be solved using the equations of motion: v = u + at and s = ut + 1/2at^2, where v is final velocity, u is initial velocity, a is acceleration, t is time, and s is displacement.

(A) To solve for the distance the automobile overtakes the truck, we need to find the time it takes for the automobile to catch up to the truck. We can use the equation v = u + at, where v is the final velocity of the automobile and u is the initial velocity of the truck (since they are traveling at the same speed, u = 9.5 m/s). We also know that the acceleration of the automobile is 2.2 m/s^2. So, plugging in these values, we get:

v = u + at
v = 9.5 + 2.2t

Next, we need to find the time it takes for the automobile to catch up to the truck. We can use the fact that at the point of overtaking, the distance traveled by both the truck and the automobile will be the same. So, we can set up an equation where the distance traveled by the truck is equal to the distance traveled by the automobile at the point of overtaking:

ut = (1/2)at^2

Substituting u = 9.5 m/s and a = 2.2 m/s^2, we get:

9.5t = (1/2)(2.2)t^2
19t = 2.2t^2

Solving for t, we get t = 8.64 seconds.

Now, we can use this value of t in the equation v = u + at to find the final velocity of the automobile:

v = 9.5 + 2.2(8.64)
v = 28.7 m/s

To find the distance traveled by the automobile, we can use the equation s = ut + 1/2at^2:

s = (9.5)(8.64) + (1/2)(2.2)(8.64)^2
s = 82.08 + 81.72
s = 163.8 meters

So, the automobile will overtake the truck at a distance of 163.8 meters beyond the traffic signal.

(B) To