How Do You Calculate Acceleration and Travel Distance from Velocity-Time Graphs?

[New2Science]

As a car passes point A on a straight road, it's speed is 10m/s, the car moves with constant acceleration, a m/s/s, along the road for T seconds until it reaches point B, where it's speed is Vm/s.
The car travels at this speed for a further 10 seconds, when it reaches the point C. From C it travels for a further T seconds with constant acceleration 3a m/s/s until it reaches a speed of 20 m/s at the point D.

Sketch the [t,v] graph for the motion, and show that V=12.5

I had no problem with showing V = 12.5 through a simultaneous equation.

However part B states,
The distance from A to D is 675m, find a and T

Any help will be so very greatly appreciated!
Thanks in advance
New2Science

 

[voko]

The area under the graph of velocity is the distance traveled. The area, as should be obvious from the sketch, is a sum of three simple areas. Together with the equation you used for velocity, you should be able to determine a and T from it.

 

[azizlwl]

If you can sketch the [t,v] graph for the motion, then you can deduce the answer.
The area under the graph is the distance.

 

[New2Science]

Thanks a lot for the help!

I got it now (:

Thanks Again

 

[Nickie]

I would approach this problem by first understanding the definitions of velocity and acceleration. Velocity is the rate of change of an object's position over time, while acceleration is the rate of change of an object's velocity over time.

Based on the given information, we know that the car starts at point A with a velocity of 10m/s and reaches a velocity of 20m/s at point D. This means that the car has undergone a change in velocity of 10m/s over a period of T seconds. Using the formula for acceleration, a = (Vf - Vi)/t, we can solve for T:

T = (20m/s - 10m/s)/a = 10/a

We also know that the car travels at a constant acceleration of 3a m/s/s for T seconds from point C to point D. Using the formula for distance, d = Vi*t + 0.5*a*t^2, we can calculate the distance traveled from point C to point D:

675m = 20m/s*T + 0.5*3a m/s/s*(T^2)

Substituting T = 10/a, we get:

675m = 20m/s*(10/a) + 0.5*3a m/s/s*((10/a)^2)

Simplifying the equation, we get:

675 = 200/a + 15/a^2

Multiplying both sides by a^2 and rearranging the equation, we get a^2 + (200/675)*a - 45 = 0

Solving for a using the quadratic formula, we get two possible values for a: a = 2.5 m/s/s or a = -18 m/s/s. However, since the car is accelerating in a positive direction, we can eliminate the negative value and conclude that a = 2.5 m/s/s.

Plugging this value back into our equation for T, we get:

T = 10/a = 10/(2.5 m/s/s) = 4 seconds

Therefore, the car accelerates at 2.5 m/s/s for 4 seconds from point A to point B, travels at a constant velocity of 20 m/s for 10 seconds from point B to point C, and accelerates at 7.5 m/s/s for 4 seconds from point C to point D.

To summarize, the