Given the acceleration as (3t^2+2)i+6e^(-2t) j+10cos 5tk, find the magnitude of velocity at t=0?

2 Answers
Apr 3, 2018

No initial points?

Explanation:

given that a =(dv)/dt rArr dv=adt rArr v=intadt + C

v_x = int a_xdt = t^3 +2t +C_1
v_y=int a_ydt = -3e^(-2t) +C_2
v_z=int a_zdt = 2sin(5t)+C_3

v_(x_t=0) = C_1
v_(y_t=0) = -3+C_2
v_(z_t=0) = C_3

and we have |v|=sqrt(v_x^2 + v_y^2 + v_z^2)

so

|v|_(t=0) = sqrt(C_1^2+(C_2-3)^2 + C_3^2)

Notice that C_1, C_2 and C_3 are constants that are calculated by your initial conditions which you didn't specify.

Apr 3, 2018

we can find the velocity in vectors by integrating the value of acceleration
i.e. vu2hati+(6/e)hatj+10sin5hatk ms^-1

Explanation:

(Given) a=(3t^2+2)hati+6e^(-2t)hatj+10cos 5thatk
so, int (3t^2+2)hati+(6e^(-2t))hatj+(10cos5hatk)
=[cancel3(t^3/cancel3)+2]hati+6(e^-(2t+1)/(-2t+1))hatj+10sin5hatk
=(t^3+2)hati+6(e^(-2t-1)/(-2t+1))hatj+10sin5hatk
now, for finding velocity put value of t=0.
=((0)^3+2)hati+6(e^(-2(0)-1)/(-2(0)+1))hatj+10sin5hatk
=2hati+6((e)^-1)/1hatj+10sin5hatk
=2hati+(6/e)hatj+10sin5hatk ms^-1