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Grade: 12th pass
        
a body starts with an initial velocity of 10m/s and is moving along a straight line with constant acceleration.when the velocity of particle is 50m/s,the acceleration is reversed in direction.the velocity of particle when it again reaches the starting point is
one year ago

Answers : (4)

Vikas TU
6887 Points
							
initial dsplacment,
50^2 = 10^2 + 2as.............(1)
After the v becomes 50,
v^2 = 50^2 – 2as............(2)
Solving both the eqns. we get,
v = 10 m/s.
Hence again he has to travel with velocity 10 m/s to reach starting point.
 
one year ago
Anant Sharma
24 Points
							V2U22aS12aS125001002aS12400When the body is decelerated,V2U22aS22aS2025002aS22500So, the body starts moving in opposite direction from that point. After coming to the initial start point (after travelling S1+S2 distance back) let the velocity of the body is V`. Therefore,V2022aS1S2V22aS12aS2V224002500V24900V4900V70msHope this information will clear your doubts about the topic.
						
10 months ago
Nishu
11 Points
							When body moves with initial velocity =10m/S and final velocity =50m/S Then, by applying v^2 - u^2 =2asWe get,                  2500-100=2as                                 1200=as......(1)When the body is returning then v=vm/SAnd u= 50m/S...... In this case acceleration as well as distance is in opposite direction therefore., they are taken as negative So v^2-u^2= 2(-a)(-s)      V^2 = 2(1200)+2500.....by(1)        V^2=4900          V= 70m/s
						
9 months ago
Niraj Bishnoi
50 Points
							
This question was raised in the Mixed-Objective type question of ETOOS. The options along with it were-
(A) 70 m/s (B) 50 m/s (C) 10 m/s (D) 49 m/s
Many people check option C) without considering the relationship between A) and D). So enough of talking let’s answer this question
u=10, v=50 in the first case
i.e v²=u²+2as
i.e 2as=2500-100=2400 ......(A)
For second case,
u=50, v=?
v²=2500+2as
=2500+2400
=4900
i.e v=sqrt(4900)
=70m/s
8 months ago
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