h(t)=t^(2)-15

If displacement S at time t is S=t^(3)-3t^(2)-15t+12 , then acceleration at time t=1 sec is

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials (iv) t^(3)-2t^(2)-15t .

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials (iv) t^(3)-2t^(2)-15t .

If S=2t^(3)-3t^(2)+15t-8 ,then the initial velocity is

Let C:r(t)=x(t)hati+y(t)hatj+z(t)hatk be a differentiable curve, i.e., lim_(xto0) (r(t+H)-r(h))/(h) exist for all t, therefore r'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk Iff r'(t) , is tangent to the curve C at the point P[x(t),y(t),z(t)] and r'(t) points in the direction of increasing t. Q. The tangent vector to r(t)=2t^(2)hati+(1-t)hatj+(3t^(2)+2)hatk at (2,0,5) is

Let C:r(t)=x(t)hati+y(t)hatj+z(t)hatk be a differentiable curve, i.e., lim_(xto0) (r(t+H)-r(h))/(h) exist for all t, therefore r'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk Iff r'(t) , is tangent to the curve C at the point P[x(t),y(t),z(t)] and r'(t) points in the direction of increasing t. Q. The tangent vector to r(t)=2t^(2)hati+(1-t)hatj+(3t^(2)+2)hatk at (2,0,5) is

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?

Differentiate the following : h(t)=(t-1/t)^(3/2)

The no of real solutions of the equation (15+sqrt(14))^(t)+(15-sqrt(14))^(t)=30 are where t=x^(2)-2|x|