A particle initially located at the origin has an acceleration of a= 5.00J^ m/s^2 and an initial velocity of v0= 8.00I^ m/s
A) Find the vector position at any time t where t is measured in seconds.
Find the vector velocity at any time t.
b) Find the coordinates of the particle at t=4.00s
x=
y=
Find the speed of the particle at this time
p(t)=p(0)+v*t + 1/2 a t^2
= 8t i + 2.50t^2 j
v(t)= v(0) + a*t= 8i+5.00 t j
To find the vector position at any time t, integrate the velocity function with respect to time.
The given initial velocity is v0 = 8.00I^ m/s. Integrating this, we get:
r(t) = ∫(8.00I^) dt = (8.00t)I^ + C
Since the particle is initially located at the origin, the constant C will be zero. Therefore, the vector position at any time t is:
r(t) = (8.00t)I^
Now, to find the vector velocity at any time t, differentiate the position function with respect to time.
v(t) = d/dt[(8.00t)I^] = 8.00I^
The vector velocity at any time t is therefore 8.00I^ m/s.
b) To find the coordinates of the particle at t = 4.00s, we can substitute t = 4.00s into the position function:
r(4) = (8.00 * 4.00)I^ = 32.00I^
Therefore, the coordinates of the particle at t = 4.00s are:
x = 32.00
y = 0
Lastly, to find the speed of the particle at this time, we can calculate the magnitude of the velocity vector at t = 4.00s:
|v(4)| = |8.00I^| = 8.00 m/s
To find the vector position at any time t, we can use the kinematic equation:
r = r0 + v0t + (1/2)at^2
where:
r is the vector position at time t,
r0 is the initial vector position (which is the origin in this case),
v0 is the initial velocity vector (8.00I^ m/s),
a is the acceleration vector (5.00J^ m/s^2), and
t is the time in seconds.
To find the vector velocity at any time t, we can use the kinematic equation:
v = v0 + at
where:
v is the vector velocity at time t,
v0 is the initial velocity vector (8.00I^ m/s),
a is the acceleration vector (5.00J^ m/s^2), and
t is the time in seconds.
Using these equations, we can solve the problems:
A) Find the vector position at any time t:
r = r0 + v0t + (1/2)at^2
= (0.00I^ + 0.00J^) + (8.00I^ m/s) * t + (1/2)(0.00I^ + 5.00J^ m/s^2) * t^2
= (8.00t)I^ + (2.50t^2)J^
Therefore, the vector position at any time t is (8.00t)I^ + (2.50t^2)J^.
B) Find the coordinates of the particle at t = 4.00s:
To find the coordinates, we simply substitute t = 4.00s into the expression for the vector position:
r = (8.00(4.00))I^ + (2.50(4.00)^2)J^
= 32.00I^ + 40.00J^
Therefore, the coordinates of the particle at t = 4.00s are x = 32.00 and y = 40.00.
To find the speed of the particle at this time, we can use the magnitude of the velocity vector, which is calculated as:
speed = |v| = sqrt(vx^2 + vy^2)
where:
|v| is the speed of the particle,
vx is the x-component of the velocity vector, and
vy is the y-component of the velocity vector.
From the equation for the vector velocity:
v = v0 + at
= (8.00I^ m/s) + (5.00J^ m/s^2) * (4.00s)
= 8.00I^ + 20.00J^
Therefore, the x-component of the velocity vector is 8.00 and the y-component is 20.00.
Using the formula for speed:
speed = sqrt(8.00^2 + 20.00^2)
= sqrt(64.00 + 400.00)
= sqrt(464.00)
≈ 21.54 m/s
Therefore, the speed of the particle at t = 4.00s is approximately 21.54 m/s.