Answer:
total time = 29.59 sec
total distance = 310.70 m
Step-by-step explanation:
given data
accelerates at a rate = 2.0 m/s²
speed = 21 m/s
slows at a constant rate = 1.1 m/s²
solution
first we take equation of motion
v = u + at ............1
21 = 0 + 2 × t
t = 10.5 sec
now in case of deceleration again we take here
0 = 21 - 1.1 × t
t = 19.09 sec
so that total time is
total time = 10.5 + 19.09
total time = 29.59 sec
and
in case of acceleration
v² = u² + 2as ...................2
21² = 0 + 2 × 2 × s
s = 110.25 m
and when it is in case of deceleration we take here
0 = 21² - 2 × 1.1 × s
s = 200.45 m
so that here total distance is
total distance = 110.25 + 200.45
total distance = 310.70 m