Answer:
acceleration = -0.042 m/s²
velocity at beginning = 14.167 m/s
velocity at end = 5.7183 m/s
Explanation:
given data
distance d1 = 1 km
distance d2 = 2 km
time t1 = 80 s
time t2 = 120 s + 80s = 200 s
to find out
acceleration and velocity at beginning and end
solution
we apply here law of motion that is
d = vt + 1/2×at²
put value
1000 = v(80) + 1/2×a(80)² ........................1
and
2000 = v(200) + 1/2×a(200)² ........................2
so from equation 1 and 2 we get a and v
a = -0.042 m/s² and
v = 14.167 m/s
so by kinematic final velocity will be
V² = v² + 2ad
V² = (14.167)² + 2×(-0.042)×(2000)
V² = 32.70
V = 5.7183 m/s
so
acceleration = -0.042 m/s²
velocity at beginning = 14.167 m/s
velocity at end = 5.7183 m/s