Find time when boy catches the girl or when they are at their closest separation..
velocity of boy is 50m/s and velocity of girl is 30m/s and acceleration of boy is 1m/s² and acceleration of girl is 2m/s² and distance between them is 150m.
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Answers
Understanding the Concept
Here the Concept of Relative Velocity and Relative Acceleration as well as Third Equation of Motion has been used. We see that we are given the velocities and acceleration of both girl and boy. Even its given the distance between the boy and the girl. And we need to find the time taken by boy to reach the nearest position of catch the girl. Here we shall take the nearest position as where the boy catches girl. Firstly we shall find the relative velocities of boy and girl. Then we shall find relative acceleration and then using these, we can find the time taken.
Let's do it !
* Formula Used :
Relative Velocity, v12 = V1
V2
This shows relative velocity of 1 with respect to 2.
Relative Acceleration, a12 = a1
a2
This shows relative velocity of 1 with respect
to 2.
圈 l 47)
= ut
+
1
2
at?
This is the Second Equation of Motion, commonly known as Position - Time Relationship.
* Solution :-
Given,
» Velocity of Boy = v, = 50 m/sec
» Velocity of Girl = V2 = 30 m/sec
» Acceleration of Boy = a, = 1 m/sec? =
» Acceleration of Girl = a, = 2 m/sec
We are already given to find that when boy catches the girl. So here all the relative motion will be Boy with respect to Girl.
For relative velocity of Boy with respect to
Girl ::
• Let the relative velocity of Boy with respect to girl be v,2
By using the formula of relative velocity,
:+ Relative Velocity, V12 = Vi
By applying values, we get
:+ Relative Velocity, V12 = =
: Relative Velocity, V12
V2
47
50
30
= 20 m sec
For relative acceleration of Boy with respect to Girl ::
• Let the relative acceleration of Boy with respect to Girl be a,2
Then by using the formula of relative acceleration, we get
Relative Acceleration, a12 = ai
By applying values, we get
150
Ak 47
40 t
2
t2
2
Transposing 2 to other side, we get
2 x 150
40 t
t?
300
40 t
t?
: t2
40t + 300
=
This forms a Quadratic Equation. Easiest way to solve this is Splitting the Middle Term. Then, we get
t2
40t +
300
This can be written as, since (30 x 10 = 300)
=
30t
10t +
300
• Taking the common terms, we get
tt
30)
10(t
30)
= 0
Now combining the common terms, we get
30)(t
10)
= 0
Here either (t - 30) = 0 or (t - 10) = 0
Then,
(t - 30) = 0 and (t - 10) = 0
t = 30 and t = 10
Clearly here we get two values of time. This means there will be two point of time when boy will be able to catch the girl which are 30 seconds and 10 seconds.
Required Time = 30 seconds or 10 secor.
* More to know:
= at V - U
This is First - Equation of Motion.
v? =
u?
2as
This is Third - Equation of Motion.
Relative Velocity, v21 = V2 - Vị
Relative Acceleration, a21
u +
a (2n
1)
= a2a1
Snth
This is the Fourth Equation of Motion.
Answer:
Maine apni
Explanation: