Question

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.​

Answer 1

Answer:

A particle starts from rest at t 0 and x 0 to move with a constant acceleration

A particle starts from rest at t = 0 and x = 0 to move with a constant acceleration = +2 ms for 20 seconds. After that, it moves with 4 m/s for the next 20 seconds. Finally, it moves with positive ! acceleration for 10 seconds until its velocity becomes zero.

Explanation:

hope it helps

Answer 2

Explanation:

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