Constant speed/acceleration

[J89]

Homework Statement

At the instant the traffic light turns green, a car that has been waiting at the intersection starts ahead with a constant acceleration of 3.20 m/s squared. At the same instant, a truck, traveling with a constant speed of 20 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?

Homework Equations

equations dealing with speed and acceleration

The Attempt at a Solution

so stuck! Don't know how to start! Help!

 

[LowlyPion]

Consider that the truck is going 20 at the moment the car starts accelerating.

At what point - like how much later - will the distance the car accelerates (x = ½*3.2*t²) be the same as the truck (x = 20*t)?

 

[thomate1]

I would approach this problem by first understanding the basic concepts of speed and acceleration. Speed is the rate at which an object moves, while acceleration is the rate at which an object's velocity (speed and direction) changes. In this scenario, the car is initially at rest and then accelerates at a constant rate of 3.20 m/s squared. The truck, on the other hand, maintains a constant speed of 20 m/s.

To answer part A of the question, we can use the equation d = v0t + 1/2at^2, where d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration. Since we are interested in the distance traveled by the car when it overtakes the truck, we can set d to be the same for both objects. The initial velocity of the car is 0 m/s, and we can assume that it overtakes the truck after t seconds. The truck has been traveling at a constant speed of 20 m/s for t seconds, so its distance traveled is 20t meters. The car, on the other hand, has been accelerating at a rate of 3.20 m/s squared for t seconds, so its distance traveled is (1/2)(3.20)(t^2) meters. Setting these two distances equal to each other, we can solve for t and then plug it back into the equation to find the distance traveled by the car.

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. We know that the car overtakes the truck when their distances traveled are equal, so we can use the same value of t that we found in part A. The initial velocity of the car is 0 m/s, and the acceleration is still 3.20 m/s squared, so we can plug in these values to find the final velocity of the car when it overtakes the truck.

In conclusion, by using basic equations for speed and acceleration, we can calculate the distance traveled and final velocity of the car when it overtakes the truck. This problem demonstrates the importance of understanding these fundamental concepts in order to solve more complex problems in physics and other scientific fields.