Answer:
The graphs are attached
Explanation:
We are told that he started at rest at rest and travelled down the hill reaching a velocity of 20 m/s after 13 seconds.
Acceleration is gotten from;
v = u + at
a = (v - u)/t
a = (20 - 0)/13
a = 20/13 m/s² or 1.54 m/s²
Distance in this period is gotten from;
v² = u² + 2as
s = (v² - u²)/2s
s = (20² - 0²)/(2 × 20/13)
s = 400/(40/13)
s = 130 m
We are told that after reaching the bottom of the hill, he travelled at a constant velocity of 20 m/s for the next 7 seconds.
At constant velocity, acceleration is 0.
Thus,distance in this period is;
s1 = vt = 20 × 7 = 140 m
I've attached the graphs