The position vector of a partcle vecr(t)=15t^(2)hati+(4-20t^(2))hatj . What is the magnitude of the acceleration (in m//s^(2) ) at t=1?
The position vector of a particle changes with time according to the relation vec(r ) (t) = 15 t^(2) hat(i) + (4 - 20 t^(2)) hat(j) What is the magnitude of the acceleration at t = 1?
If v=(t^(2)-4t+10^(5)) m/s where t is in second. Find acceleration at t=1 sec.
A particle moves in circle of radius 1.0 cm at a speed given by v=2.0 t where v is in cm/s and t in seconeds. A. find the radia accelerationof the particle at t=1 s. b. Findthe tangential acceleration at t=1s. c.Find the magnitude of the aceleration at t=1s.
A particle moves in such a way that its position vector at any time t is vec(r)=that(i)+1/2 t^(2)hat(j)+that(k) . Find as a function of time: (i) The velocity ((dvec(r))/(dt)) (ii) The speed (|(dvec(r))/(dt)|) (iii) The acceleration ((dvec(v))/(dt)) (iv) The magnitude of the acceleration (v) The magnitude of the component of acceleration along velocity (called tangential acceleration) (v) The magnitude of the component of acceleration perpendicular to velocity (called normal acceleration).
If the displacement of a particle is (2t^(2) +t +5) meter then, what will be acceleration at t = 5sec.
The displacement s of a moving particle at a time t is given by s=5+20t-2t^(2) . Find its acceleration when the velocity is zero.
If v=(t+2)(t+3) then acceleration (i.e(dv)/(dt)) at t=1 sec.
If displacement S at time t is S=t^(3)-3t^(2)-15t+12 , then acceleration at time t=1 sec is
The magnetic flux in a closed circuit of resistance 20 Omega , varies with time (t) according to equation phi=8t^(2)-6t+5 . What is the magnitude of induced current at time t=1 sec ?
