I'm stuck on a.
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You have to estimate the slope of the tangent line to the graph at t = 10 s. To do that, you can use points on the graph very close to t = 10 s, essentially applying the mean value theorem.

The MVT says that for some time t between two fixed instances a and b, one can guarantee that the slope of the secant line through (a, v(a) ) and (b, v(b) ) is equal to the slope of the tangent line through t. In this case, this would be saying that the instantaneous acceleration at t = 10 s is approximately equal to the average acceleration over some interval surrounding t = 10 s. The smaller the interval, the better the approximation.

For instance, the plot suggests that the velocity at t = 9 s is nearly 45 m/s, while the velocity at t = 11 s is nearly 47 m/s. Then the average acceleration over this interval is

(47 m/s - 45 m/s) / (11 s - 9 s) = (2 m/s) / (2 s) = 1 m/s²

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation:

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