Another motion problem for constant acceleration

In summary, a truck traveling at a constant speed of 28.0 m/s approaches a stationary car that is 125 m ahead. When the car starts moving with an acceleration of 2.6 m/s2, the truck will take 6.32 seconds to reach the car. This can be solved using the position as a function of time for both the truck and the car, setting them equal, and using the fact that the truck moves at a constant speed while the car accelerates.
  • #1
chroncile
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Homework Statement


A truck travels at a constant speed of 28.0 m/s in the fast lane of a two-lane highway. It approaches a stationary car stopped a the side of the road. When the truck is still 125 m behind the car, the car pulls out into the slow lane with an acceleration of 2.6 m/s2. How long will it take the truck to reach the car? (Answer is 6.32 s)


Homework Equations


N / A


The Attempt at a Solution


I have no idea how to do this :(
 
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  • #2
This is similar to the wolf/rabbit problem.

Write the position as a function of time for both the truck and the car. Measure from the initial position of the truck when the car starts moving. Set them equal.
 
  • #3
I'm sorry, but could you please write that out in equation form? I am not a clever man.
 
  • #4
Oh, you're clever enough. :smile:

You just solved an almost identical problem! Hint: The truck moves at constant speed; the car accelerates.
 
  • #5


I would approach this problem by first identifying the known values and variables. The known values in this problem are the truck's constant speed of 28.0 m/s, the distance between the truck and the car (125 m), and the car's acceleration of 2.6 m/s2. The variable we are trying to solve for is time (t).

Next, I would use the kinematic equation for constant acceleration, which is d = vit + 1/2at^2, where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. We can rearrange this equation to solve for t, which gives us t = (-vi ± √(vi^2 + 2ad)) / a.

Plugging in the known values, we get t = (-28.0 ± √(28.0^2 + 2(2.6)(125))) / 2.6. This gives us two possible values for t, but we can disregard the negative value since time cannot be negative. Therefore, the truck will take 6.32 seconds to reach the car.

In conclusion, by using the kinematic equation for constant acceleration, we can determine the time it will take for the truck to reach the car after the car pulls out into the slow lane. This type of problem is commonly used in physics to calculate the motion of objects with constant acceleration.
 

FAQ: Another motion problem for constant acceleration

1. What is constant acceleration?

Constant acceleration is a type of motion in which the velocity of an object changes by the same amount in every unit of time. This can be represented by a straight line on a velocity-time graph, with the slope of the line representing the constant acceleration.

2. How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. This formula assumes that the acceleration is constant throughout the motion.

3. What is the difference between constant acceleration and uniform motion?

Uniform motion is a type of motion in which the velocity of an object remains constant over time, while constant acceleration involves a changing velocity. In uniform motion, the velocity-time graph is a straight line, while in constant acceleration, the graph is a curved line.

4. How does air resistance affect constant acceleration?

Air resistance can affect constant acceleration by slowing down the object's motion and causing a decrease in acceleration. This is because air resistance creates a force in the opposite direction of the object's motion, which must be overcome by the object's acceleration.

5. What are some real-life examples of constant acceleration?

Some real-life examples of constant acceleration include a car accelerating from a stop sign, a rollercoaster going downhill, a ball rolling down a ramp, and a rocket launching into space. In all of these cases, the acceleration remains constant throughout the motion.

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