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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

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

Grade:12th pass

6 Answers

Vikas TU
14149 Points
6 years ago
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.
 
Anant Sharma
24 Points
5 years ago
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.
Nishu
11 Points
5 years ago
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
Niraj Bishnoi
50 Points
5 years ago
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
Hemashree46
13 Points
4 years ago
Given:
First,
u=10 m/s
v=50m/s
  V^2 =u^2 +2as
50^2=10^2+2as
2as =2500-100
2as =2400----1
Second,
u=50m/s
  v^2=u^2+2as
 V^2=50^2+2as
2as=v^2-2500----2
From 1 and 2
V^2-2500=2400
V^2=4900
V=√4900=70
    V=70m/s
 
 
Kushagra Madhukar
askIITians Faculty 628 Points
2 years ago
Dear student,
Please find the answer to your 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=√(4900)
=70m/s
 
Thanks and regards,
Kushagra

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