~~~~Mechanics question~~~~~

1.
Consider the first interval (time = 3 s):
u = ?
a = ?
t = 3 s
s = 153 m
s = ut + (1/2)at2
153 = u(3) + (1/2)a(3)2
3a + 2u = 102 ...... (1)

Consider both the two intervals (time = 3+5 = 8 s):
u = ?
a = ?
t = 8 s
s = 153 + 215 = 368 m
s = ut + (1/2)at2
368 = u(8) + (1/2)a(8)2
4a + u = 46 ...... (2)

2x(2)-(1): 5a = -10
Hence, a = -2 m s-2

4x(1)-3x(2): 5u = 270
Hence, u = 54 m s-1

Consider that the object comes to rest.
u = 54 m s-1
v = 0 m s-1
a = -2 m s-2
s = ?
v2 = u2 + 2as
(o) = (54)2 + 2(-2)s
4s = 2916
s = 729 m

Ans: a) The acceleration = -2 m s-2 (deceleration)
Ans: b) The initial speed = 54 m s-1
Ans: c) The distance traveled when the body comes to rest = 729 m

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2.
Let d m be the distance of the uphill/downhill trip.

Time taken for the uphill trip = d/2 m s-1

Time taken for the downhill trip = d/3 m s-1

Total distance of the round trip = 2d m

Total time taken of the round trip = d/2 + d/3 = 5d/6 s

Average speed of the round trip = 2d/(5d/6) = 2.4 m s-1
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