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A particle starts from rest it, s acceleration at an instant t is a=10/v+1 m/s where v is instantaneous velocity. the distance traveled after which the particle attain a velocity of 15 m/s?
Verified Answer
A particle starts from rest it, s acceleration at an instant t is a=10...
a = v dv/ds
 
Now
10 / (v+1) = v dv/ds
 
Now
 
10 ds = v (v+1) dv
 
on integrating
 
10 s = v^3/3  + v^2/ 2  + c
 
at v=0
 
s=0
 
hence
10 s = v^3/3  + v^2/ 2  
 
 
now if v = 15 m/sec
 
10s = 3375/3 + 225/2
 
10s = 1125 + 112.5
 
10s = 1237.5
 
s = 123.75 m
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Most Upvoted Answer
A particle starts from rest it, s acceleration at an instant t is a=10...
Given information:
- Initial velocity (u) = 0 m/s (particle starts from rest)
- Acceleration (a) at an instant t = 10/v1 m/s (where v1 is the instantaneous velocity)

To find:
The distance traveled when the particle attains a velocity of 15 m/s.

Approach:
To solve this problem, we can use the kinematic equation that relates velocity, acceleration, and displacement:

v^2 = u^2 + 2as

Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the displacement

We need to find the displacement s when the velocity v is 15 m/s.

Solution:

Step 1: Find the acceleration at the instant when the velocity is 15 m/s.

Given: a = 10/v1
At v = 15 m/s, substitute the value:
a = 10/15 = 2/3 m/s^2

Step 2: Use the kinematic equation to find the displacement s.

v^2 = u^2 + 2as

Given: u = 0 m/s, a = 2/3 m/s^2, v = 15 m/s

(15)^2 = (0)^2 + 2 * (2/3) * s

225 = 4/3 * s

225 * 3/4 = s

s = 168.75 m

Step 3: Conclusion

The particle will travel a distance of 168.75 meters when it attains a velocity of 15 m/s.
Community Answer
A particle starts from rest it, s acceleration at an instant t is a=10...
11.25
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