Member advised to use the homework template for posts in the homework sections of PF.
I tried but I can't.
Write dv/dt in terms of x and y, as defined in this question.First I used the formula dv/dt but I am used to have a ƒ(x) in the questions I usually did.
Sorry I disagree. I think that's deliberate.There's a conflict in the figure, t2 is stated as being 3.8, but shows as 4.0 in the figure.
I wasn't paying attention to the limited choices of possible answers, so the accuracy doesn't need to be that great. There's no information after t = 4.0, so what's being calculated is the average acceleration from t = 3.8 to t = 4.0, not the instantaneous acceleration at t = 4.0. Based on the graph and the data, the acceleration is decreasing with time, instantaneous acceleration at t = 4.0 would be less.It states the velocity at t=3.8 seconds is 130m/s and asks you to calculate the acceleration at t2=4.0 seconds. The graph gives you enough data to calculate the velocity at t2=4.0. So you have the velocities at two points a known time apart.
The accuracy might not be great but you can calculate an answer.