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. vu#2hati+(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#