A partical of mass 0.4 kg is moving with a constant speed of 10 m per secand towards north is subjected to a constant force of 8 ndirected towards south for 30 secand suppose that the force is applied at t=0 and the particle at the that time is at x=0 predict the position of the particle at t=0sec ,25sec,100secend
Answers
Explanation:
I can' t understand it is so long
Answer:
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 ms-2
(i) At t = –5 s
Acceleration, a' = 0 and u = 10 m/s
s = ut + (1/2) a' t2
= 10 × (–5) = –50 m
(ii) At t = 25 s
Acceleration, a'' = –20 m/s2 and u = 10 m/s
s' = ut' + (1/2) a" t2
= 10 × 25 + (1/2) × (-20) × (25)2
= 250 - 6250 = -6000 m
(iii) At t = 100 s
For 0 ≤ t ≤ 30 s
a = -20 ms-2
u = 10 m/s
s1 = ut + (1/2)a"t2
= 10 × 30 + (1/2) × (-20) × (30)2
= 300 - 9000 = -8700 m
For 30 < t ≤ 100 s
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 + (1/2) a" t2
= -590 × 70 = -41300 m
∴ Total distance, s" = s1 + s2 = -8700 -41300 = -50000 m = -50 km.
Explanation: