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Question
A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
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Solution
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = -8.0 N
Acceleration produced in the body, `a = "F"/"m" = (-8.0)/(0.40) = -20 "m/s"^2`
(i) At t = -5s
Acceleration, a' = 0 and u = 10 m/s
`"S" = "ut" + 1/2 "at"^2`
= 10 × (-5) = -50 m
(ii) At t = 25 s
Acceleration, a'' = -20 m/s2 and u = 10 m/s
S = ut + `1/2` at2
`= 10 xx 25 + 1/2 xx (-20) xx (25)^2`
= 250 + 6250 = -6000 m
(iii) At t = 100 s
For `0 <= t <= 30`
a = –20 m/s2
u = 10 m/s
S1 = ut + `1/2` at2
`= 10 xx 30 + 1/2 xx (-20) xx (30)^2`
= 300 - 9000
= -8700 m
For 30' < t <= 100s
As per the first equation of motion, for t = 30 s, final velocity is given as:
v = u + at
= 10 + (-20) × 30 = -590 m/s
Velocity of the body after 30 s = -590 m/s
For motion between 30 s to 100 s, i.e., in 70 s:
S2 = vt
= -590 × 70 = -41300 m
∴ Total distance, S = s1 + s2 = -8700 - 41300 = -50000 m
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