Motion Along A Straight Line - HELP

In summary, the car starts with a constant acceleration of 3.15 at the instant the traffic light turns green and overtakes a truck traveling with a constant speed of 18.0 at the same instant. The equations of motion for both vehicles can be used to find the time and distance when they meet. However, without additional information, it is not possible to determine how far beyond its starting point the car overtakes the truck.
  • #1
Arika
2
0
At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.15 . At the same instant a truck, traveling with a constant speed of 18.0 , overtakes and passes the car.

Part A: How far beyond its starting point does the car overtake the truck?
Part B: How fast is the car traveling when it overtakes the truck?



Vo of the car is zero and the acc. = 3.15
Vo of the truck is 18 and the acc. is zero,
i need atleast a hint,i don't get part A,i don't have enough information:S But i think part B would be easy once i know part A
 
Physics news on Phys.org
  • #2
Basically, you need to find the equations of motion for both vehicles. It means the dependence of coordinate upon time. In your case it has the form

[tex]x(t) = At^2 + Bt + C[/tex]

and velocity

[tex]v(t) = 2At + B[/tex]

Just fix the constants A, B, and C to meet the boundary conditions. Once you know the constants for both vehicles, you can equate their positions to get the time(s) when they meet and then calculate the point of meeting, as well as any other quantity of interest.
 
  • #3
I'm sorry but I don't get it, from where do i bring A,B and C?

I think the only equations we can use are:

[tex]V = Vo + at[/tex]
[tex]V^2 = Vo^2 + 2aX[/tex]
[tex]X = Vot + .5 at^2[/tex]
[tex]X = Vt - .5 at^2[/tex]
[tex]X = .5 (Vo+V) t[/tex]

V being final velocity, Vo initial velocity, t time, a acceleration,X displacement.
 

1. What is motion along a straight line?

Motion along a straight line refers to the movement of an object in a straight path, without deviating or changing direction.

2. What are the different types of motion along a straight line?

There are three main types of motion along a straight line: uniform motion, accelerated motion, and decelerated motion. Uniform motion is when an object moves at a constant speed in a straight line. Accelerated motion is when an object increases its speed in a straight line. Decelerated motion is when an object decreases its speed in a straight line.

3. How is motion along a straight line measured?

Motion along a straight line is measured using a variety of tools such as rulers, tape measures, and motion sensors. The distance an object travels in a straight line is typically measured in meters (m), and the time it takes to travel that distance is measured in seconds (s).

4. What factors can affect motion along a straight line?

There are several factors that can affect motion along a straight line, including the force applied to the object, the mass of the object, and any external forces such as friction or air resistance. Additionally, the shape and surface of the object can also impact its motion along a straight line.

5. How is motion along a straight line represented graphically?

Motion along a straight line can be represented graphically using a distance-time graph or a velocity-time graph. In a distance-time graph, the distance traveled by the object is plotted on the y-axis and time is plotted on the x-axis. In a velocity-time graph, the velocity (speed and direction) of the object is plotted on the y-axis and time is plotted on the x-axis.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
727
  • Advanced Physics Homework Help
Replies
19
Views
1K
  • Electromagnetism
Replies
3
Views
1K
  • Introductory Physics Homework Help
3
Replies
98
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • General Math
Replies
4
Views
1K
  • General Math
Replies
2
Views
782
  • Advanced Physics Homework Help
Replies
3
Views
3K
Replies
10
Views
2K
  • General Math
Replies
6
Views
695
Back
Top