Velocity time graph find position, acce?

[fashion_fever]

The velocity-versus-time graph is shown for a particle moving along the x-axis. Its initial position is xo=4.80 m at to=0.00 s. What is the particle's position at t=1.25 s if the total time the particle moves is 5.00 s and the maximum velocity is 2.00 m/s?


1)At t=3.75 s, what is the particle's position?

2)At t=3.75 s, what is the particle's acceleration?

I tried to find the position by finding the area up until 3.75,but the answer seems not right...
and acceleration i used v final - v initial / change in time. but I am not sure if it's right or not..

if some1 can help with the steps, that'll be great~thanks

 

[CharliH]

Acceleration is the slope of a velocity-time graph.
Change in position is the area under the graph from the start time to the end time. - Remember to add this to the initial position

Until you do calculus, you pretty much need to learn that by heart. (If you have some calculus, the above basically says a = dv/dt and that change in position is the integral of velocity.)

Slope = change in y/change in x
You need to calculate this for a straight portion of the graph. If there are kinks in the graph, it means the acceleration is changing. You need to work on a part where it's constant.

 

[fashion_fever]

got it thanks.

 

[sawan9695]

how would you find the area for last small triangle 3-3.75s
1/5(0.75)(what would be height ?

 

[rwisz]

1) To find the particle's position at t=1.25 s, we can use the formula for position on a velocity-time graph: x = xo + vot + 1/2at^2. Plugging in the given values, we get x = 4.80 + (2.00)(1.25) + 1/2(0)(1.25)^2 = 7.55 m. Therefore, the particle's position at t=1.25 s is 7.55 m.

2) To find the particle's acceleration at t=3.75 s, we can use the formula for acceleration on a velocity-time graph: a = (vf - vi)/t. Plugging in the given values, we get a = (2.00 - 0)/(3.75) = 0.53 m/s^2. Therefore, the particle's acceleration at t=3.75 s is 0.53 m/s^2.