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When you have information about the motion of the cars in terms of its velocity of acceleration over a period of time but not its distance from a fixed object.

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Q: When you use a graph to solve a problem about how far a car traveled?
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When using a graph to solve a problem about how far a car traveled during a specified time during which it was accelerating how many area calculations do you have to make and what are the shapes?

The answer depends on the variables plotted in the graph.


When you use a graph to solve a problem about how far a car traveled during a specified time during which it was accelerating how many area calculations do you have to make?

To calculate the distance traveled from a velocity-time graph, you need to calculate the area under the graph. The number of calculations you need to make, and the shapes you divide the area into will depend on the shape of the curve.The skinnier you make the slices, and the more area measurements you make,the more accurate your answer will be.In the kind of math called "calculus", there's a way to work that problem as if thewidth of the slices was zero and there was an infinite number of them, so that the answer comes out exactly right. It's called "integration."


What trend can you see in the graph you made from the data table about the distance traveled by a car?

That question can only be answered by the person who made the graph from the data table, referred to as "you" in the question. Get busy!


What is the speed of a car whose distance time graph indicated that it traveled 200m in 25 s?

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What is a linear equation that fits this situation A car is traveling at a constant velocity After two hours the car has traveled 90 miles After five hours the car has traveled 225 miles?

With constant velocity (v) the equation for distance can be d = vt, where d is the distance traveled (in miles), and t is the time (in hours). So at t= 2 hr, d = 90 mi, solve for v = (90 mi)/(2 hr) = 45 mi/hr (or mph). Check this at 5 hours: d = (45 mi/hr)*(5 hr) = 225 mi, which checks with the original problem.


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