Overtaking Truck: How Far & How Fast for Car?

[aznriceboiz5]

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.55 . At the same instant a truck, traveling with a constant speed of 21.0 , overtakes and passes the car.


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


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


Thanks for the help

 

[curly_ebhc]

Ok you have two equations of motion 1 for constant velocity and one for constant acceleration.

Now you have two options.
Solve both equations for displacement/distance and make two data sets. I suggest graphing both data sets on a shared set of axes. Make time the hotizontal axis and position the vertical axis. This will give you the best sense of what is actually happening. Don't forget that the cars position will change as a smooth curve. This is the more time involved method.

Second method:
Use your eqns. Don't forget that the vehicles will pass when they have the same position (in this case displacement).

This is where I leave you for now. Good luck.

 

[quicknote]

!

I would approach this scenario by first defining the variables involved. We have the acceleration of the car, which is 3.55 m/s^2, the initial speed of the truck, which is 21.0 m/s, and the distance and speed of the car at the moment it overtakes the truck.

Using the equations of motion, we can calculate the distance traveled by the car and the speed of the car at the moment it overtakes the truck. The distance traveled by the car can be calculated using the equation d = v0t + 1/2at^2, where d is the distance, v0 is the initial speed, a is the acceleration, and t is the time. In this case, the initial speed of the car is 0 m/s, as it starts from rest, and the time can be calculated using the equation t = (vf - v0)/a, where vf is the final speed and v0 is the initial speed.

Plugging in the values, we get t = (21.0 m/s - 0 m/s)/3.55 m/s^2 = 5.92 s. Using this value of time in the equation for distance, we get d = (0 m/s)(5.92 s) + 1/2(3.55 m/s^2)(5.92 s)^2 = 62.9 m. Therefore, the car overtakes the truck at a distance of 62.9 m beyond its starting point.

To calculate the speed of the car at the moment it overtakes the truck, we can use the equation vf = v0 + at, where vf is the final speed, v0 is the initial speed, a is the acceleration, and t is the time. In this case, we know the acceleration and the time, and the initial speed is still 0 m/s. Plugging in the values, we get vf = (0 m/s) + (3.55 m/s^2)(5.92 s) = 21.0 m/s. Therefore, the car is traveling at a speed of 21.0 m/s when it overtakes the truck.

It is important to note that these calculations are based on ideal conditions and do not take into account factors such as air resistance, friction, and the potential differences in acceleration and speed of the car and truck. Further experimentation or more detailed