Position versus time graph help
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Position versus time graph help

[From: ] [author: ] [Date: 12-09-01] [Hit: ]
but the y-axis is labeled velocity.If the graph is velocity vs time, then the velocity is uniformly increasing (in the forward direction for the top curve, inthe reverse direction for the bottom curve).However, if the title is correct,......
Please take a look at the pic of the graph below

http://img841.imageshack.us/img841/2829/photo1ikm.jpg

Please help me with the following questions

a) Describe,in words, the velocity of each cars. Make sure you discuss each car's speed and direction.
b) Calculate the velocity of each of the cars

Help with these 2 questions will be much appreciated

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The graph is confusing. The title says position vs time, but the y-axis is labeled velocity. If the graph is velocity vs time, then the velocity is uniformly increasing (in the forward direction for the top curve, in the reverse direction for the bottom curve). However, if the title is correct, and the y axis is position, then the velocity is the slope of the curve and is constant for both curves. For the top curve it would be 2 m/s forward and for the bottom it would be 2 m/s in reverse. (+2 m/s and -2 m/s)

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Line going up to the right - the car is moving with uniform acceleration.

Line going to the left - to avoid any confusion I will just say the car has uniform negative acceleration.

also calculate the velocity of each of the cars when? At what time constant?

If you are referring to average velocity find the area of the section underneath the graph, so find the (area of the trapezium for car 1) and (triangle for car 2), which will give you the displacement, then divide by the change in time.

Area of trapezium = (sum of parallel sides * perpendicular height)/2
then you just plug in your values

This is ONLY APPLICABLE if this is a velocity vs time graph, if it isn't then you are going to have to find the gradient to find the velocities
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