Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.
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Answer:
3.69 m/s
Explanation:
Forces :
mgsin Θ - mumgcosΘ = ma
g x sinΘ - mu x g x cosΘ = a
9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a
a = -1.337 m/s²
so you have final velocity = 0 m/s
initial velocity = ? m/s
Given d = 5.1 m
By kinematics
vf² = vo² + 2ad
0 = vo² + 2 x -1.337*5.1
vo = 3.69 m/s
v = average speed of movement of the Southwest Indian Ridge = 20 mm/year
d = distance moved by the Southwest Indian Ridge = 100 mm
t = number of years required to move distance "d"
distance traveled is given as
d = v t
inserting the above values in the formula
100 mm = (20 mm/year) t
dividing both side by 20 mm/year
t = 100 mm/(20 mm/year)
t = 5 years
