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A particle moves along a street line with an initial velocity of 5 m per second.If it experiences a constant acceleration of 3 m per second.How much distress will it cover in the food?Second of its motion?
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A particle moves along a street line with an initial velocity of 5 m p...
Initial Conditions
- Initial velocity (u) = 5 m/s
- Constant acceleration (a) = 3 m/s²
Distance Covered in the First Second
To find the distance covered in the first second of motion, we can use the equation of motion:
- Distance (s) = u*t + (1/2)*a*t²
Where:
- t = time (1 second for the first interval)
Substituting the values:
- s = 5*(1) + (1/2)*3*(1)²
- s = 5 + (1.5)
- s = 6.5 meters
Total Distance Covered in the First Two Seconds
To find the total distance covered in the first two seconds, we again use the same equation of motion for t = 2 seconds:
- s = u*t + (1/2)*a*t²
Substituting the values:
- s = 5*(2) + (1/2)*3*(2)²
- s = 10 + (1/2)*3*4
- s = 10 + 6
- s = 16 meters
Summary
- Distance covered in the first second: 6.5 meters
- Total distance covered in the first two seconds: 16 meters
This analysis helps us understand how the initial velocity and constant acceleration affect the distance traveled by the particle over time.
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A particle moves along a street line with an initial velocity of 5 m per second.If it experiences a constant acceleration of 3 m per second.How much distress will it cover in the food?Second of its motion?
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