Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details
Get extra R 2,800 off
USE CODE: chait6

```				   A particle moving in a straight line covers half the distance with speed of 3m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is?
```

5 years ago

Share

```										Let total distance=d
Total time=T
Then in  d/2.......
t1=d/6

t2=d/(4*4.5)  +  d/(4*7.5)
=4d/45

T=t1+t2=d/6+4d/45 =23d/90

Average speed =t. Dis/t. Time= d/T =3.91
m/s........ANS
All the best!!!!!

```
5 years ago
```										let total distance of path is S ...

for first half , time taken if t
speed = distance/time
3 = S/2t
t = S/6      .............1

for second half , time taken is T ...
for first  T/2 it moves with 4.5
speed = distance/time
4.5 = 2x/T
x = 4.5T/2   ...........2
for second T/2 it moves with 7.5
7.5 = 2y/T
y = 7.5T/2       ..........3
we have , x+y = S/2so
(7.5+4.5)T = S/2
T = S/24        ...............4
total time taken during its motion is T+t
Ttotal = T+t = S/24+S/6 = 5S/24
total distance = S
average speed = total distance/total time = S/5S/24 = 24/5 = 4.8m/s

approve if u like my ans
```
5 years ago
```										hi
let the distanc covered with speed 3m/s be x.mts
time taken is x/3sec.
ii)second half(x mts)
time taken to travel second half will be 18x/11.25 sec
avg. speed=total distanc /totaltime
avg.speed=1.9m/s
plzz. prove!!!!!!!!!!!!!!!!!!!!!!!!!!
```
5 years ago
```										Let total distance be S. Total time be T and V1= 3m/s be the velocity for which it covers the first half of the distance and V2 = 4.5 m/s and V3 = 7.5 m/s be the velocities in the second half of the distance. Let <v> be the average velocity that has to be calculated.
Let the other half be completed in times t1 and t2 (t1= t2=t, from the question).
For Ist half of the distance:
S/2= V1(T-2t)
=> T = S/2V1 + 2t              ....(1)
For 2nd half of the distance:
S/2 = (V2 + V3)2t
=> S = 4(V2 + V3)t             ....(2)
From (1) and (2).
T = {2(V2 + V3)/V1    + 2}t      ....(3)
we know that,
<v> = total distance/total time
so,
<v> = S/T
Putting the values of  T and S from (3) and (2),
<v> = 4(V2 + V2)t /{2(V2 + V3)/V1   +2 }t
=> <v> = 2V1 (V2 + V3)/(2V1 + V2 + V3)
putting the values of V1, V2 and V3.
<v> = 4 m/s.
```
5 years ago
```										all the answers are waste of time there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s
```
one month ago
```										all the answers are waste of time there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s answered by a 8th grade
```
one month ago

# Other Related Questions on Mechanics

a packet is released from a balloon accelerating upward with acceleration a. the acceleration of packet just after the release is

Since the ballon was moving upward with acceleration a, there is a force of gravity on the packet too. The balloon is moving upward bexause the reaction force R>W, weight of the packet. As...

 Shaswata Biswas 3 months ago

Just after the release the packet posses only the component of the velocity aquired. Here the accleration of the balloon has no effect on the acceration of the body. When the body is...

 Shaswata Biswas 3 months ago

The accleration of the packet will be `g` acting downward as no external force other then gravity working on it.it will achieve the case of free fall. If I am wrong please ..explain the...

 Chandan kumar mandal 2 months ago
The potential energy of a particle of mass 1kg moving along x-axis is given by U(x)=[x^2/2-x]J. If total mechanical energy of the particle is 2J find its maximum speed.

total mechanical energy = U+ K.E to attain maximum speed the obect must have maximum K.E as K.E will be maximum , U has to be minimum (conservation of energy )given - U(x) = x²/2-xfor this...

 fizaparveen 2 months ago

We know that the object accelerate till force is applied on it and it attains maximum velocity just after the force becomes zero... SO we know the negative of potential energy gradient is...

 Suraj Singh 20 days ago

Ans- 2Speed will be max when kinetic energy is maximum so potential energy will be minimum so differentiating the function put it equal to 0 so we obtain minimum for x=0 now kin potential...

 rishabh doshi 2 months ago
if a body move with u velocity then after collide with wall and remove with v velocity what is the impulse

Hi aman, I think this might help as we know that Impulse and also F=m(dv/dt) so (here m is the mass of body and dv/dt =acc) here p is the momentum ans delta p is change in momentum hence in...

 Ankit Jaiswal 3 months ago
what is work energy thereom...................................?

The work-energy theorem is a generalized description of motion which states that work done by the sum of all forces acting on an object is equal to change in that object’s kinetic energy....

 dolly bhatia one month ago

The work W done by the net force on a particle equals the change in the particle`s kinetic energy KE: W = Δ K E = 1 2 m v f 2 − 1 2 m v i 2 . The work-energy theorem can be derived from...

 Sushmit Trivedi one month ago

For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation...

 rahul kushwaha 3 months ago
The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM

Let the angle of throwing be Q with respect to the ground So, the range is R = 40*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...

 Tapas Khanda 5 months ago

Let the angle of throwing be Q with respect to the ground So, the range is R = 30*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...

 Tapas Khanda 5 months ago

Diagram is not required to solve this question. You can directly use the formula to find range : R = (v^2/g)*sin2Q Maximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8) So,...

 Tapas Khanda 5 months ago
What harmfull gases are produced when we burn plastics? And can we burn plastics?

I`m doing a project, can someone help me, How to reduce those toxic contents when plantic are burnt Like sulphur, nitrogen etc

 2017 years ago
View all Questions »

• Complete Physics Course - Class 12
• OFFERED PRICE: R 2,600
• View Details
Get extra R 2,600 off
USE CODE: chait6

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details

Get extra R 2,800 off
USE CODE: chait6

More Questions On Mechanics