The applicable relationship here is
... acceleration = rate of change of velocity
4. The slope of the velocity curve is constant, so the acceleration is constant. That slope is 3 m/s in 5 s, or (3 m/s)/(5 s) = (3/5) m/s² = 0.6 m/s²
7. You're looking for a point on the velocity curve where its slope is -2 m/s². That will be somewhere between t=0 and t=4, because slope is positive for t>4. The only available choice in that region is t = 2 s.
8. At 6.00 m/s² for 3 seconds, velocity will change (3 s)×(6.00 m/s²) = 18.00 m/s. That would get you from an initial speed of 3.44 m/s to 21.44 m/s, so clearly 3 s is almost right, but a little too long for the given change in velocity. The best choice is 2.91 s.
If you want to actually figure it out, the change in velocity is 20.9 -3.44 = 17.46 m/s. Using the relation a = ∆v/∆t, we can rearrange it to ∆t = ∆v/a, or
... ∆t = (17.46 m/s)/(6 m/s²) = 2.91 s
9. This problem combines the determination of acceleration with a units conversion problem. Numbers in the problem are given in kph and seconds, and answers are given in m/s². At some point, you need to convert from km/h to m/s. The multiplier for that is (1000 m/km)/(3600 s/h) = 1/3.6 (m·h)/(km·s).
Your change in speed is -24.6 km/h = (1/3.6)·(-24.6) m/s ≈ -6.8333 m/s. When that change in speed occurs over 3.56 seconds, the acceleration is
... (-6.8333 m/s)/(3.56 s) ≈ -1.919 m/s² ≈ -1.92 m/s²
5. At 10 s, the velocity is about 14 m/s. Looking for grid points the curve goes through, we can use (11, 16) and (9, 12). That is, over the 2-second range from 9 s to 11 s, the velocity increases 4 m/s from 12 m/s to 16 m/s. The acceleration is
... ∆v/∆t = a = (4 m/s)/(2 s) = 2 m/s²
6. ∆v/∆t = a = (28 m/s - 0 m/s)/(4.22 s) ≈ 6.6351 m/s² ≈ 6.64 m/s²
10. No change in velocity means the acceleration is 0 m/s².
