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A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance traveled in next 2 sec-?
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A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming ...
Given:
Initial velocity (u) = 0
Distance covered in first 5 sec (s1) = 10 m
Time taken to cover s1 (t1) = 5 sec
Distance covered in next 3 sec (s2) = 10 m
Time taken to cover s2 (t2) = 3 sec
Time for next 2 sec (t3) = 2 sec

Calculations:
Total distance covered (s) = s1 + s2
s = 10 + 10 = 20 m

Let's assume the acceleration of the particle is a.

Using the first equation of motion, we get:
s1 = ut1 + 1/2 at1^2
10 = 0*5 + 1/2*a*5^2
a = 2 m/s^2

Now, using the third equation of motion, we can find the distance covered in the next 2 seconds:
s3 = ut3 + 1/2 at3^2
s3 = 0*2 + 1/2*2*2^2
s3 = 4 m

Therefore, the distance traveled in the next 2 seconds is 4 m.

Conclusion:
The particle will travel a distance of 4 m in the next 2 seconds assuming constant acceleration.
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A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming ...
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A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance traveled in next 2 sec-?
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