3.1 Motion

Question 1

What is the distance between a velocity-time graph and a speed-time graph?

Answer 1

1. Velocity is a vector, speed is a scalar.
2. Therefore, velocity-time graphs can have a negative part to show that something is travelling in an opposite direction;

Question 2

What does the gradient of a v-t graph give, and why?

Answer 2

Acceleration; because the gradient is the change in y over the change in x. In this case, this is v/t, which is acceleration.

Question 3

Why is acceleration a vector?

Answer 3

Because the equation for acceleration is v/t, which involves velocity, which is a scalar quantity. Therefore, acceleration must have a direction as well as a magnitude, which is the definition of a vector.

Question 4

What does a linear v-t graph tell you?

Answer 4

1. That there is a constant gradient, and so uniform acceleration;
2. That SUVAT can be used, because these equations only work with linear acceleration
3. This means that in this case, the graph in the form y=mx+c will have the same equation as v=u+at
4. Therefore, u will be = c, which means that the initial velocity is the same as the graph’s y axis intercept;

Question 5

How should one find displacement from a v-t graph? Why should they want to find displacement, and not distance?

Answer 5

1. The area under the graph;
2. This is because the distance travelled = average speed x time;
3. This is displacement and not distance, because displacement is a vector quantity, which the v-t graph deals with, because a and v are all vector quantities;

Question 6

What does a negative area under a v-t graph mean?

Answer 6

That the object is moving back to its starting point.

Question 7

What are the two ways that non-uniform acceleration can be shown on a v-t graph?

Answer 7

1. Increasing acceleration:
   This is shown by an increasing gradient, that mirrors y=x^2 graphs;
2. Decreasing acceleration:
   This is shown by an decreasing gradient, that mirrors y=-x^2 graphs;
   In both cases, the acceleration is non-linear, and so the graphs will not be straight. This means that SUVAT cannot be used, and therefore the equation of the graph cannot be expressed in the form of v=u+at.

Question 8

Which equations should one find the displacement of a v-t graph? List the equations that should be used in the eventuality that the acceleration is uniform and non uniform.

Answer 8

The displacement of a v-t graph is the same as the area underneath it:
1. If the graph is linear (uniform acceleration), the only equation that should be used is the area of a triangle: 1/2 base x height;
2. If the graph is non-linear (non-uniform acceleration), then the area under the graph should be split into trapeziums. To find the area of a trapezium;
1/2 (a+b)h. The area under the graph, of the triangle that is closest to the origin will also need to be used, by finding 1/2 bh. The total area under the graph will be the sum of all of the areas of these trapeziums.