Need help with this problem on position and velocity. I'm given a graph of velocity vs. time, how do I draw the acceleration vs. time from these graph?(question 2) And then I am given acceleration vs. time and need to draw velocity vs. time from these graph (question 3). Please see attachment.© BrainMass Inc. brainmass.com December 24, 2021, 7:38 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
Let's start by defining the key terms:
• velocity is the rate at which an object changes its position
• acceleration is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.
Looking at the definitions you can see that velocity and acceleration are related and it is this connectivity which will allow you to develop graphs of acceleration vs time using velocity vs time graphs and vice versa. To think more about this connection consider a car moving with a constant velocity of +10 m/s. Even though the car is moving it is not accelerating. A car moving with a constant velocity has an acceleration of 0 m/s^2. Now consider a car moving with a changing velocity. A car with a changing velocity will have an acceleration.
Now let's think about what the graphs are telling us. To start with velocity vs time: the shape of a velocity versus time graph reveals pertinent information about an object's acceleration. A line sloping up from left to right tells us that the car has a positive velocity (is getting faster) and since the velocity is changing the object has an acceleration. If the velocity vs time line is horizontal then velocity is constant or not changing - and if velocity is not changing then acceleration is zero. If velocity is negative (slowing) the velocity vs time graph will have a line sloping down from left to right and acceleration will be negative. Acceleration is directly related to the slope of the line on the velocity vs time scale. To calculate acceleration from velocity find the slope of the line. To find slope divide rise (vertical difference) by run (horizontal difference).
Lets go through your first graph as an example:
There are four different sections to consider in this graph -
1) A positive velocity (+slope)
2) A constant velocity (horizontal line)
3) A negative velocity (-slope)
4) Another positive velocity (+slope)
For each change in velocity you need to calculate an acceleration by calculating the slope.
1) The first 1 second of the graph is a positive slope. In that second the velocity changes from 0 m/s to 0.5 m/s. To calculate slope use this equation: velocity time 2 - velocity time 1/time 2 - time 1 or in this case 0.5 - 0/1-0 (0.5 minus zero divided by one minus zero - * do the subtraction and then the division). So the slope is 0.5 and the acceleration is 0.5 m/s^2 (meters per second squared).
2) For the next second, time 1 second to 2 seconds, the line is horizontal which means a slope of zero and an acceleration of zero (or no acceleration)
3) For the next two seconds (time 2 sec to 4 sec) velocity changes from 0.5 m/s to -1m/s. So slope equals -1 - 0.5/4-2 which equals -0.75 m/s^2.
4) For the last two seconds (time 4 sec. to 6 secs.) velocity changes from 0 m/s to 1 m/s. So slope equals 1-0/6-4 or 0.5 m/s^2.
*Note that acceleration vs time graphs only have straight horizontal or vertical lines not diagonal.
To convert an acceleration vs time graph to a velocity vs time graph you need to examine the lines on the acceleration graph as they tell us the slope to draw on the velocity graph.
Let's look at your first acceleration graph as example.
In this graph there are five different rates of acceleration.
1) Time 0 to time 2 seconds, acceleration is zero so the velocity slope will be zero and velocity will be constant. Since your instructions say that velocity at time zero is 0 m/s then this velocity is maintained. Time 0 to time 2 velocity is 0 m/s.
2) Time 2 to time 3 seconds, acceleration is 1 m/s^2. So the velocity graph will have a positively line with slope 1. Remember slope is velocity 2 - velocity 1/time 2-time 1. To figure out velocity you need to do a bit of algebra (or just think about it as these values are rather simple). Plugging in the values we know 1 m/s^2 = v2-0/1, now solve for v2. v2 equals 1. So your line should be a diagonal from 0m/s at time 2 to 1 m/s at time 3.
3) Time 3 to time 4 seconds, acceleration is 0.5; velocity slope will be positive 0.5.
4) Time 4 to time 5 seconds, acceleration and velocity slope will be -0.5
5) Time 5 to time 6 seconds acceleration and velocity slope will be zero.
You may also find this website helpful: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/1DKin/U1L4b.html© BrainMass Inc. brainmass.com December 24, 2021, 7:38 pm ad1c9bdddf>