CBSE Class 11-science Answered
The particle is moving on a straight line path with initial velocity 45m/s and along +ve x-axis and acceleration of 10. along -ve x-axis. The distance travelled by the particle in 5th second of its motion is ?
Asked by arunavamitra50 | 21 Jun, 2018, 07:05: AM
Expert Answer
Inititial velocity u0 = 45 m/s, acceleration a = -10 m/s2
Let u5 be initial velocity for 5th second travel. u5 = 45-10×4 = 5 m/s . ( formula used " v = u+at " )
distance S5 travelled in 5th second, S5 = 5×1 +(1/2)×(-10)×1×1 = 0.
(formula used for calculating S5, " S = ut+(1/2)at2 " with t = 1 because it is one second travel)
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