Perhaps they want you to draw a graph of V(t), with a smooth line through the points, and measure the slope of the tangent to that curve at t = 9s.
The acceleration rate is not constant for your set of numbers, so there will be some uncertainty based upon how you draw the curve.
A very good estimate of the acceleration at t = 9s can be obtained by dividing the velocity change between t = 8 and 10 s by the time interval (2 s)..
a(t=9) = (15 - 10)/2 = 7.5 m/s^2
Time() 0 2 4 6 8 10 12 14 16
Velocity() 0 0 2 5 10 15 20 22 22
By measuring the slope of your curve, find the magnitude of the instantaneous acceleration at times = 9s
i don't know how to solve it
4 answers
i did the graph already and i put the anwser 7.5m/s^2 and i got it wrong
i got it thank
Sorry my mistake. Right approach; math/typing error. I meant 2.5 m/s^2
4 answers