Given a graph find aceleration

[Alexanddros81]

Homework Statement

Serway Physics Section 2.4 Acceleration

17. Figure P2.17 shows a graph of $v_x$ versus t for the motion of a
motorcyclist as he starts from rest and moves along the road in a straight
line. (a) Find the average acceleration for the time interval t = 0 to t = 6.00 s.
(b) Estimate the time at which the acceleration has its greatest positive
value and the value of the acceleration at that instant. (c) When is the
acceleration zero? (d) Estimate the maximum negative value of the acceleration
and the time at which it occurs.

Homework Equations

The Attempt at a Solution

At part (b) i don't know how to find the value of the acceleration at t = 3s.
Any hints?

 

[mjc123]

Estimate the slope of the graph at 3s. Consider the points at 2s and 4s.

 

[Alexanddros81]

So I calculate the average acceleration for t = 2s to t = 4s?

$a_{avg} = \frac {v_{xf} - v_{xi}} {t_{f} - t_{i}} = \frac {6 m/s - 2 m/s} {4s - 2s} = \frac {4m/s} {2s} = 2m/s^2$

Is this correct?

 

[mjc123]

Looks like a good estimate. The actual acceleration at 3s will be a bit greater, but not much, to judge from the graph.

 

[Alexanddros81]

So for part (d) the maximum negative acceleration is at t=8s
and the value is given by $a_{avg} = \frac {v_{xf} - v_{xi}} {t_{f} - t_{i}} = \frac {5 m/s - 8 m/s} {9s - 7s} = \frac {-3m/s} {2s} = -1.5m/s^2$
Is this correct?

 

[mjc123]

Not bad approximation, I think.