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A particle is moving along a straight line such that its position depends on time as x= 1-at bt^2. where a = 2 m/s, b = 1 m/s². The distance covered by the particle during the first 3 seconds from start of the motion will be:?
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A particle is moving along a straight line such that its position depe...
Explanation:
The given equation of motion is x= 1-at bt^2, where a = 2 m/s, b = 1 m/s².
Step 1: Find the position of the particle at t=0
x= 1-a(0) b(0)^2
x= 1
Therefore, the particle starts at x=1.
Step 2: Find the velocity of the particle using the derivative of x with respect to t
v= dx/dt = -a-2bt
At t=0, v= -a= -2 m/s.
The negative sign indicates that the particle is moving in the negative direction.
Step 3: Find the distance covered by the particle during the first 3 seconds.
The distance covered by the particle can be calculated using the area under the velocity-time graph.
From t=0 to t=3, the velocity is constant and equal to -a.
Therefore, the distance covered by the particle during the first 3 seconds is:
distance = velocity x time = -a x 3 = -6 m
The negative sign indicates that the particle has moved in the negative direction.
Step 4: Find the final position of the particle at t=3
x= 1-a(3) b(3)^2
x= -16
Therefore, the final position of the particle at t=3 is x=-16.
Conclusion:
The particle starts at x=1 and moves in the negative direction with a constant velocity of -2 m/s. During the first 3 seconds, the particle covers a distance of -6 m and reaches the final position of x=-16.
The given equation of motion is x= 1-at bt^2, where a = 2 m/s, b = 1 m/s².
Step 1: Find the position of the particle at t=0
x= 1-a(0) b(0)^2
x= 1
Therefore, the particle starts at x=1.
Step 2: Find the velocity of the particle using the derivative of x with respect to t
v= dx/dt = -a-2bt
At t=0, v= -a= -2 m/s.
The negative sign indicates that the particle is moving in the negative direction.
Step 3: Find the distance covered by the particle during the first 3 seconds.
The distance covered by the particle can be calculated using the area under the velocity-time graph.
From t=0 to t=3, the velocity is constant and equal to -a.
Therefore, the distance covered by the particle during the first 3 seconds is:
distance = velocity x time = -a x 3 = -6 m
The negative sign indicates that the particle has moved in the negative direction.
Step 4: Find the final position of the particle at t=3
x= 1-a(3) b(3)^2
x= -16
Therefore, the final position of the particle at t=3 is x=-16.
Conclusion:
The particle starts at x=1 and moves in the negative direction with a constant velocity of -2 m/s. During the first 3 seconds, the particle covers a distance of -6 m and reaches the final position of x=-16.
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A particle is moving along a straight line such that its position depends on time as x= 1-at bt^2. where a = 2 m/s, b = 1 m/s². The distance covered by the particle during the first 3 seconds from start of the motion will be:?
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