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ICSE Class 9 Answered

A particle moves the distance of 20 m in 5th second and distance of 8 m in 7th second of its motion .What is total distance moved by the particle ? Its answer is 188 m please tell me how?
Asked by ayushmani1871 | 26 Jul, 2019, 11:05: PM
answered-by-expert Expert Answer
Let u be the initial speed and a be its uniform acceleration.
 
after 4 seconds, speed of particle is obtained from equation " v = u+at ", 
 
Hence speed of particle after 4 seconds =  u+a(4) = u+4a 
 
Distance S5 travelled in 5th second is obtained from the equation S = u t + (1/2) a t2  by considering t = 1 s
 
hence Distance S5 travelled in 5th second is given by,   S5 = (u+4a) + (1/2)a  = u + (9/2)a  = 20 m     ............................(1)
 
Hence speed of particle after 6 seconds =  u+a(6) = u+6a 
 
hence Distance S7 travelled in 7th second is given by,   S7 = (u+6a) + (1/2)a  = u + (13/2)a  = 8 m     ............................(2)
 
By solving eqn.(1) and eqn.(2), we get u = 47 m/s and  a = -6 m/s2
 
since acceleration is negative, it is retarded motion.
 
Distance S travelled by particle, before coming to rest is calculated using equation  v2 = u2 - 2aS by considering final speed v = 0
 
Hence, Distance S travelled by particle, before coming to rest is given by,  S = u2/(2a)  = (47×47)/(2×6) ≈ 184 m
Answered by Thiyagarajan K | 27 Jul, 2019, 11:24: AM

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