# Motion along a straight line (Finding acceleration).

• Sarah Sposato
In summary: You should revisit the problem and try to analyze it more carefully before jumping into calculations. Also, make sure you use the correct formulas for the given information.
Sarah Sposato
Figure 2-24 shows a red car and a green car that move toward each other. Figure 2-25 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35 m at time t = 0. The green car has a constant speed of -20 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
The graph of their motion shows the two cars intersect their position at t = 12 seconds.

I started out by interpreting the labels and information.
Xo (g) : 270m
Xo (r) : -35m
(Constant) V (g) : -20m/s
Vo (r) : 0 m/s.
t = 12 seconds.

1. I used the formula X = Xo + ( V - Vo/ 2 )t to get the position at which the cars meet.

X = 270 + (-20/2)12

X = 150m,

Then I used the formula X = Xo + Vot + 1/2at^2 to get the acceleration of the red car.

150 = -35 + 0 (12) + 1/2a(12)^2.

This yielded a = 1.68 m/s*sThe answer is supposed to be a= 0.9m/s*s.Where did I go wrong? Was my interpretation of the data incorrect? Or just the way I plugged in the data?

Can you post the two figures associated with this problem?

Sarah Sposato said:
Figure 2-24 shows a red car and a green car that move toward each other. Figure 2-25 is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35 m at time t = 0. The green car has a constant speed of -20 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?
The graph of their motion shows the two cars intersect their position at t = 12 seconds.

I started out by interpreting the labels and information.
Xo (g) : 270m
Xo (r) : -35m
(Constant) V (g) : -20m/s
Vo (r) : 0 m/s.
t = 12 seconds.

1. I used the formula X = Xo + ( V - Vo/ 2 )t to get the position at which the cars meet.
Why? The green car has a constant speed of -20 m/s, according to the problem statement.

What's the correct formula for finding distance traveled for an object traveling at constant speed?

X = 270 + (-20/2)12

X = 150m,
This distance X doesn't appear to be correct, just by looking at the graph which shows the motion of the cars versus time.

Then I used the formula X = Xo + Vot + 1/2at^2 to get the acceleration of the red car.

150 = -35 + 0 (12) + 1/2a(12)^2.

This yielded a = 1.68 m/s*s

The answer is supposed to be a= 0.9m/s*s.

Where did I go wrong? Was my interpretation of the data incorrect? Or just the way I plugged in the data?
I think the second part of your work went wrong because you made the mistakes in the first part, as discussed above.

## What is motion along a straight line?

Motion along a straight line refers to the movement of an object in a single direction, either forward or backward, without any change in direction.

## What is acceleration?

Acceleration is the rate of change of an object's velocity with respect to time. It is a measure of how quickly an object is speeding up or slowing down.

## How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (v2 - v1) / t, where a is acceleration, v2 is the final velocity, v1 is the initial velocity, and t is the time interval.

## What are the units of acceleration?

The units of acceleration are typically meters per second squared (m/s^2) in the metric system and feet per second squared (ft/s^2) in the imperial system.

## What factors can affect acceleration?

Acceleration can be affected by factors such as the force acting on an object, the mass of the object, and the direction of the force relative to the motion of the object.

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