How do tou find the acceleration of a car between two points on a velocity-time variation graph

A = ∆v/∆t
Say velocity is on the y-axis and time in on the x-axis. If you take the points (2,1) and (5,8) for example, the difference between the velocity points is 8-1 = 7. This difference is the change in velocity (i.e. ∆v, ∆ means change in) between these two points. The same goes for time: 5-2 = 3.
So we've concluded that in this example, ∆v = 7 and ∆t = 3. Now put these into the equation above to find acceleration:
7/3 = 2.333...
So the acceleration between these two points on the graph is 2.333...(recurring), or as a fraction, 7/3.

P = 928.09 Pa

Explanation:

Given that,

Mass of a gas cylinder, m = 90 kg

The diameter of the cylinder, d = 1.1 m

Radius, r = 0.55 m

We need to find the ground pressure. The pressure is equal to force per unit area. So,

So, the ground pressure is 928.09 Pa.