Question

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

Answer 1

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.

Answer 2

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`