Solve Kinematics Problem: Car Overtaking Truck

[fxsnowy]

I need help with this problem. Any help is greatly appreciated.

Homework Statement

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.40m/s^2 . At the same instant a truck, traveling with a constant speed of 17.5m/s , overtakes and passes the car.

Homework Equations

How far beyond its starting point does the car overtake the truck?

How fast is the car traveling when it overtakes the truck?

The Attempt at a Solution

 

[Darth Frodo]

What do you think should be done? How would you go about solving the problem? HAve you any attempt made?

 

[fxsnowy]

What do you think should be done? How would you go about solving the problem? HAve you any attempt made?

There is no specific time given, so i tried to use the formula v^2f = v^2i + 2a(xf-xi).
But for the car, I can't solve for the final velocity since I don't have the final position x so I don't know how to go about solving this.

 

[Darth Frodo]

May I ask if you have the answer to question 1? I have an answer but I want to ensure it's correct.

Also ask yourself, at the instant the car overtakes the truck, what is common between the two so I can equate something?

 

[Nickie]

To solve this problem, we can use the kinematics equations:

1. For the first part, we can use the equation x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial position for both the car and the truck is 0, since they both start at the same point. The car's acceleration is 3.40m/s^2 and the truck's speed is 17.5m/s.

Plugging in these values, we get:

x = 0 + 0 + 1/2(3.40)t^2 for the car

x = 0 + 17.5t for the truck

Since we are looking for the distance at which the car overtakes the truck, we can set these two equations equal to each other:

0 + 0 + 1/2(3.40)t^2 = 0 + 17.5t

Solving for t, we get t = 5.15 seconds.

To find the distance, we can plug this value back into either equation. Using the car's equation, we get:

x = 0 + 0 + 1/2(3.40)(5.15)^2 = 44.7m

Therefore, the car overtakes the truck at a distance of 44.7m from the intersection.

2. For the second part, 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, we are looking for the car's velocity when it overtakes the truck, so we can use the same value for t as before, 5.15 seconds.

Plugging in the values, we get:

v = 0 + (3.40)(5.15) = 17.5m/s

Therefore, the car is traveling at the same speed as the truck, 17.5m/s, when it overtakes the truck.

In conclusion, the car overtakes the truck at a distance of 44.7m from the intersection and at a speed of 17.5m/s.