The vector position of a particle varies in time according to the expression r⃗ =3.00i^−6.00t2j^ where r is in meters and t is in seconds.
Determine the velocity of the particle at t = 2.0 s.
To find the velocity of the particle at t = 2.0 s, we need to take the derivative of the position vector with respect to time.
Given: r⃗ = 3.00i^ - 6.00t^2j^
Step 1: Take the derivative of r with respect to t:
d(r⃗)/dt = d(3.00i^ - 6.00t^2j^)/dt
Step 2: Take the derivative of each component separately:
dx/dt = d(3.00i^)/dt = 0
dy/dt = d(-6.00t^2j^)/dt = -12.00t
Step 3: Substitute t = 2.0 s into the derivative equation:
dy/dt = -12.00(2.0) = -24.00 m/s
Therefore, the velocity of the particle at t = 2.0 s is -24.00 m/s in the y-direction.
To determine the velocity of the particle at t = 2.0 s, we need to take the derivative of the position vector with respect to time. The position vector is given by r⃗ = 3.00i^ - 6.00t^2j^.
Taking the derivative of r⃗, we get:
dr⃗ / dt = d/dt(3.00i^ - 6.00t^2j^).
To find the derivative, we apply the power rule of differentiation, which states that the derivative of t^n with respect to t is n*t^(n-1).
Differentiating the i^ component:
d/dt(3.00i^) = 0, since i^ does not depend on time.
Differentiating the j^ component:
d/dt(-6.00t^2j^) = (-6.00)*(d/dt(t^2))*j^.
Applying the power rule, d/dt(t^2) = 2t.
Therefore, d/dt(-6.00t^2j^) = (-6.00)*(2t)*j^.
Combining the components, we get the velocity vector v⃗:
v⃗ = 0i^ - 12.00tj^.
To find the velocity at t = 2.0 s, substitute t = 2.0 into the velocity equation:
v⃗ = 0i^ - 12.00(2.0)j^.
Simplifying the equation:
v⃗ = 0i^ - 24.00j^.
Therefore, the velocity of the particle at t = 2.0 s is 0i^ - 24.00j^ m/s.
v(t) = 0i - 12j
v(2) = -24j