A particle starts from rest at t = 0 and x = 0 to move with a constant acceleration = +2 m/s2, for 20 seconds. After that, it moves with 4 m/s2 for the next 20 seconds. Finally, it moves with positive acceleration for 10 seconds until its velocity becomes zero.
(a) What is the value of the acceleration in the last phase of motion?
(b) What is the final x-coordinate of the particle?
(c) Find the total distance covered by the particle during the whole motion.
Answers
Answer:
Explanation:
Correction in question :
Finally it moves with negative acceleration for 10 seconds until its velocity becomes zero
u = 0 m/s at t = 0 & x = 0
a = 2m/s² for 20 sec
V = u + at = 0 + 2 * 20 = 40 m/s
S = ut + (1/2)at² = 0 + (1/2)2 * 20² = 400 m
t = 20 , v = 40 m/s x = 400
a = 4m/s² for next 20 sec
V = 40 + 4(20) = 120 m/s
S = 40*20 + (1/2)4 * 20² = 1600 m
at t = 40 v = 120 m/s x = 400+ 1600 = 2000
0 = 120 + a(10)
=> a = -12 m/s²
S = 120 * 10 + (1/2)(-12)10² = 1200 - 600 = 600m
at t = 50 v = 0 X = 2000 + 600 = 2600
Final x coordinate = 2600
Total Distance covered = 2600 m
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Now rephrase question as second acceleration is -ve and keeping positive acceleration for 10 seconds
u = 0 m/s at t = 0 & x = 0
a = 2m/s² for 20 sec
V = u + at = 0 + 2 * 20 = 40 m/s
S = ut + (1/2)at² = 0 + (1/2)2 * 20² = 400 m
t = 20 , v = 40 m/s x = 400
a = -4m/s² for next 20 sec
V = 40 - 4(20) = -40 m/s
S = 40*20 + (1/2)(-4) * 20² = -800 m
at t = 40 v = -40 m/s x = 400 -800 = -400
0 = -40 + a(10)
=> a = 4 m/s²
S = (-40) * 10 + (1/2)(4)10² = -400 + 200 = -200 m
at t = 50 v = 0 X = -400 -200 = -600
Final x coordinate = -600
Total Distance covered = 400 + 800 + 200 = 1400 m