CBSE Class 11-science Answered
A body with uniform acceleration covers 76 m in 5th second and 116 m in 10th second. (a)Find the distance traveled in 15th second. (b) What is the distance traveled in next 5 seconds?
Asked by Mãnï kåûr | 02 Aug, 2019, 05:09: PM
Expert Answer
Let u is initial velocity and a is acceleration for the given motion
initial velocity for nth second travel is obtained from the formula " v = u+at " by considering time duration t =1
Distance travelled at nth second is obtained from the formula, " S = ut+(1/2)at2 " by considering appropriate initial velocity and time duration t = 1
For the 5th second travel, initial velocity u+4a, hence distance S5 travelled in 5th second ,
S5 = ( u+4a )+ (1/2)a = 76 or u + (9/2)a = 76 m ..................... (1)
For the 10th second travel, initial velocity u+9a, hence distance S10 travelled in 10th second ,
S10 = ( u+9a )+ (1/2)a = 116 or u + (19/2)a = 116 m..................... (2)
By solving eqn.(1) and eqn.(2), we get u = 40 m/s and a = 8 m/s2
Distance S15 travelled in 15th second ,
S15 = ( u+14a )+ (1/2)a = [40+ (14×8)]+(1/2)×8 = 156 m
Distance S travelled in next 5 seconds, S = (u+15a)×t +(1/2)at2 = [40+(15×8)]×5 + (1/2)×8×25 = 900 m
Answered by Thiyagarajan K | 03 Aug, 2019, 09:47: AM
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