The object accelerates for the first 4 s (0 to 4), then moves at a constant speed for the next 4 s (4 to 8), then speeds up again for the next 2 s (8 to 10), moves at another constant speed for 6 s (10 to 16), then slows to a stop for the last 4 s (16 to 20).
In the times where the speed is linearly increasing, the acceleration is constant and equal to the slope of those line segments. Where the speed is constant, acceleration is 0.
Recall that average acceleration is given by
<em>a</em> = ∆<em>v</em> / ∆<em>t</em>
So we get
[i] <em>a</em> = (20 m/s - 0) / (4 s - 0) = 5 m/s²
[ii] <em>a</em> = 0 (because the speed is constant at 12 s)
[iii] Not really sure what is meant by "total acceleration" here... Is it the sum of all accelerations observed here? If so, we know the acceleration 5 m/s² on [0, 4]; it's 0 for both [4, 8] and [10, 16]; and we can compute the acceleration for the remaining two intervals to be
[8, 10]: <em>a</em> = (30 m/s - 20 m/s) / (10 s - 8 s) = 5 m/s²
[16, 20]: <em>a</em> = (0 m/s - 30 m/s) / (20 s - 16 s) = -7.5 m/s²
Then the total <em>might</em> be 2.5 m/s². <em>Maybe</em>.
[iv] We can compute distance traveled by finding the area under the velocity curve. The area under the curve for the first 10 s can be split up into a triangle on [0, 2] with height 20 m/s and base 4 s; a rectangle on [4, 8] with height 20 m/s and base 4 s; and a trapezoid on [8, 10] with "base" lengths 20 m/s and 30 m/s, and "height" 2s. The total area is then
1/2 (20 m/s) (4 s) + (20 m/s) (4 s) + 1/2 (20 m/s + 30 m/s) (2 s) = 170 m
[v] We already have the distance traveled over the first 10 s, so just subtract the area of the trapezoid:
1/2 (20 m/s) (4 s) + (20 m/s) (4 s) = 120 m
[vi] This would be the acceleration we found in [iii] over the interval [16, 20]:
<em>a</em> = (0 m/s - 30 m/s) / (20 s - 16 s) = -7.5 m/s²
[vii] According to the plot, the velocity at 16 s is 30 m/s.
[viii] According to the plot, at 2s the velocity is 10 m/s.
