A particle moves along the x −
axis in such a way that its acceleration at time
t
for
t
>
0
is given
a
(
t
)
=
3
t
2
. When
t
=
1
, the position of the particle is
6
and the velocity is
2.

(a) Write an equation for the velocity,
ν
(
t
)
, of the particle for all
t
>
0
.

(b) Write an equation for the position,
x
(
t
)
, of the particle for all
t
>
0
.

(c) Find the position of the particle when
t
=
e
. Give an exact answer.

It looks like the particle's acceleration is

a(t) = 3t²

and we're given the initial velocity and position to be v(1) = 2 and x(1) = 6.

(a) By the fundamental theorem of calculus, the velocity at time t is

v(t) = v(1) + ∫₁ᵗ a(u) du

v(t) = 2 + ∫₁ᵗ 3u² du

v(t) = 2 + u³ |₁ᵗ

v(t) = 2 + (t³ - 1³)

v(t) = t³ + 1

(b) Use the theorem again to get the position at time t,

x(t) = x(1) + ∫₁ᵗ v(u) du

x(t) = 6 + ∫₁ᵗ (u³ + 1) du

x(t) = 6 + (1/4 u⁴ + u) |₁ᵗ

x(t) = 6 + (1/4 t⁴ + t - 1/4 • 1⁴ - 1)

x(t) = 1/4 t⁴ + t + 19/4

(c) The position of the particle at t = e is

x(e) = 1/4 e⁴ + e + 19/4

sorry not sure

step-by-step explanation:

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)