Calculate the distance travelled by the van between 16.0s and 23.0s. give your answer correct to 2 significant figures.

45.5 meters
Looking at the graph, the van is decelerating at a constant rate from 13 m/s down to 0 m/s over the time interval from 16.0s to 23.0s. The duration of the deceleration is (23.0 - 16.0) = 7.0s. The change in velocity over that interval is 13m/s, so the rate of acceleration is (13.0/7) = -1.857143 m/s^2 (Value is negative since the vehicle is decelerating.
Since d=1/2 A T^2 for constant acceleration, the formula for how far the van travels is
13T - 1/2*1.857143 T^2. = d
Substituting 7 for T gives
13*7 - 1/2*1.857143*7^2 = d
91 - 0.928571*49 = d
91 - 45.5 = d
45.5 = d
Therefore the van moves a total of 45.5 meters over the time interval from 16.0s to 23.0s.

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.