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

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

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Answer 1 · 2018-04-03T16:25:49

No initial points?

Explanation:

given that $a = \frac{d v}{d t} \Rightarrow d v = a d t \Rightarrow v = \int a d t + C$ $a =(dv)/dt rArr dv=adt rArr v=intadt + C$

$v_{x} = \int a_{x} d t = t^{3} + 2 t + C_{1}$ $v_x = int a_xdt = t^3 +2t +C_1$
$v_{y} = \int a_{y} d t = - 3 e^{- 2 t} + C_{2}$ $v_y=int a_ydt = -3e^(-2t) +C_2$
$v_{z} = \int a_{z} d t = 2 sin (5 t) + C_{3}$ $v_z=int a_zdt = 2sin(5t)+C_3$

$v_{x_{t} = 0} = C_{1}$ $v_(x_t=0) = C_1$
$v_{y_{t} = 0} = - 3 + C_{2}$ $v_(y_t=0) = -3+C_2$
$v_{z_{t} = 0} = C_{3}$ $v_(z_t=0) = C_3$

and we have $| v | = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$ $|v|=sqrt(v_x^2 + v_y^2 + v_z^2)$

so

${| v |}_{t = 0} = \sqrt{C_{1}^{2} + {(C_{2} - 3)}_{2}^{2} + C_{3}^{2}}$ $|v|_(t=0) = sqrt(C_1^2+(C_2-3)^2 + C_3^2)$

Notice that $C_{1}, C_{2}$ $C_1, C_2$ and $C_{3}$ $C_3$ are constants that are calculated by your initial conditions which you didn't specify.

Answer 2 · 2018-04-03T17:07:42

we can find the velocity in vectors by integrating the value of acceleration
i.e. vu $2 ˆ i + (\frac{6}{e}) ˆ j + 10 sin 5 ˆ k$ $2hati+(6/e)hatj+10sin5hatk$ $m s^{- 1}$ $ms^-1$

Explanation:

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

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

2 Answers

Explanation:

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