On a speed time graph the slope measurses acceleration?

Answer 1

If an object is allowed to free-fall under gravity, it will accelerate downwards at 9.8 metres per second per second (m s-2). This is equal to the standard gravity, symbol g, which on the surface of Earth has a value of around 9.8 newtons per kilogram (N kg-1) or 9.8 m s-2. Both units are equivalent.

This is because the acceleration of an object is given by the net force that acts on it (in newtons) divided by its mass (in kilograms). This is Newton's second law:

F = m a

So a = F / m

In the case of a free-falling object, the only force acting on the object is gravity, and the magnitude of this force is the mass of the object multiplied by standard gravity.

F = m g

So the acceleration a = g, the standard gravity.

On the other hand, consider an object at rest on a horizontal surface. The force of gravity still acts on the object, but it is being counteracted by an upward force provided by the surface, which is equal and opposite to the force of gravity.

An object on a slope is somewhere between these two. Part of the force of gravity is counteracted by an upward force provided by the slope, but the remainder causes the object to accelerate down the slope (assuming the object is round or slippery). In fact we need to resolve the force into its components by using trigonometry.

To answer the question, if the slope is at an angle theta (θ) to the horizontal, then the acceleration is given by:

a = g sin(θ)

If the slope is perfectly horizontal then θ = 0, and sin (0) = 0, so a = 0, and the object does not move.

Answer 2

You can find acceleration by calculating the slope of a time vs. velocity graph. Time can be measured in seconds, and velocity in meters per second. The slope, change in y over change in x, would yield the units of m/s2, the units of acceleration.

Answer 3

The graph of velocity vs time.

Not speed vs time. Acceleration is a vector as is velocity, but speed is not. When going round a circular path at constant speed, your velocity is constantly changing (its direction) and you are always accelerating (towards the centre of the circular path).

Answer 4

Yes, you can analyse acceleration with a graph of speed vs. time. Or for a more sophisticated analysis, you could use differential calculus.

Answer 5

Yes it does. Better call it instantaneous acceleration.

Answer 6

Rate of change in the acceleration.