Position of a particle moving along the straight line given by y=3t3 +t2+5
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The given question is incomplete. Please add this part in the question.
Find the velocity and acceleration at t = 2 s.
Solution:
Here, y = 3 t^3 + t^2 +5
Here position y is measured in m.
For velocity, we will differentiate above equation with respect to t.
Velocity (v) = dy /dt = 9 t^2 +2 t
Now, we shall find velocity at t=2 s,
v = 9 *(2)^2 + 2 * 2 = 40 m/s
For acceleration, we will differentiate above velocity equation with respect to t.
Acceleration (a) = dv / dt = 18 t +2
Now, we shall find acceleration at t = 2 s
a = 18 * 2 + 2 = 38 m / s^2
Find the velocity and acceleration at t = 2 s.
Solution:
Here, y = 3 t^3 + t^2 +5
Here position y is measured in m.
For velocity, we will differentiate above equation with respect to t.
Velocity (v) = dy /dt = 9 t^2 +2 t
Now, we shall find velocity at t=2 s,
v = 9 *(2)^2 + 2 * 2 = 40 m/s
For acceleration, we will differentiate above velocity equation with respect to t.
Acceleration (a) = dv / dt = 18 t +2
Now, we shall find acceleration at t = 2 s
a = 18 * 2 + 2 = 38 m / s^2
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