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The acceleration 'a' in m/s² of a particle is given by a=3t^2 2t 2 where t is the time. If the particle starts out with a velocity, u=2m/s at t=0, then the velocity at the end of 2 seconds will be?
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The acceleration 'a' in m/s² of a particle is given by a=3t^2 2t 2 whe...
Explanation:
Given, acceleration (a) = 3t^2 + 2t + 2 m/s²
Initial velocity (u) = 2 m/s
Time (t) = 2 s
We need to find the final velocity (v) of the particle after 2 seconds.
Step 1: Finding the velocity of the particle at any time (t)
We know that acceleration (a) is the rate of change of velocity (v) with respect to time (t).
So, integrating the acceleration with respect to time, we get velocity (v) at any time (t).
v = ∫a dt = ∫(3t^2 + 2t + 2) dt = t^3 + t^2 + 2t + C
where C is the constant of integration.
Step 2: Finding the constant of integration (C)
We can find the value of C using the initial velocity (u).
At t = 0, v = u
So, u = 0^3 + 0^2 + 2(0) + C = C
Therefore, C = 2 m/s.
Step 3: Finding the final velocity (v) after 2 seconds
Now, putting the values of C and t in the equation of velocity, we get
v = t^3 + t^2 + 2t + 2 m/s
At t = 2 s,
v = 2^3 + 2^2 + 2(2) + 2 = 18 m/s
Therefore, the final velocity (v) of the particle after 2 seconds is 18 m/s.
Given, acceleration (a) = 3t^2 + 2t + 2 m/s²
Initial velocity (u) = 2 m/s
Time (t) = 2 s
We need to find the final velocity (v) of the particle after 2 seconds.
Step 1: Finding the velocity of the particle at any time (t)
We know that acceleration (a) is the rate of change of velocity (v) with respect to time (t).
So, integrating the acceleration with respect to time, we get velocity (v) at any time (t).
v = ∫a dt = ∫(3t^2 + 2t + 2) dt = t^3 + t^2 + 2t + C
where C is the constant of integration.
Step 2: Finding the constant of integration (C)
We can find the value of C using the initial velocity (u).
At t = 0, v = u
So, u = 0^3 + 0^2 + 2(0) + C = C
Therefore, C = 2 m/s.
Step 3: Finding the final velocity (v) after 2 seconds
Now, putting the values of C and t in the equation of velocity, we get
v = t^3 + t^2 + 2t + 2 m/s
At t = 2 s,
v = 2^3 + 2^2 + 2(2) + 2 = 18 m/s
Therefore, the final velocity (v) of the particle after 2 seconds is 18 m/s.
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The acceleration 'a' in m/s² of a particle is given by a=3t^2 2t 2 where t is the time. If the particle starts out with a velocity, u=2m/s at t=0, then the velocity at the end of 2 seconds will be?
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