the acceleration of a particle is defined by the equation a= 4-t^2 with an initial velocity of two millimeters per second. Find the exact equation for the position of the particle if its initial position was zero.
if a = 4 - t^2 , then
v = 4t - (1/3)t^3 + c
given: when t=0, v = 2 mm/s
2 = 0 - 0 + c
c = 2
v = 4t - (1/3)t^2 + 2
s = 2t^2 - (1/12)t^4 + 2t + k
when t=0 , s = 0 , so k = 0
s = 2t^2 - (1/12)t^4 + 2t
check by taking the derivative twice
x"(t)=d²x/da²=a=4-t^2
Integrate two times and substitute the initial conditions to find the required equation:
x'(t)
=∫a dt
=∫(4-t^2)dt
=4t-(t^3/3)+C1
From x'(t)=2 mm-t^-1, we get
2 mm-t^-1 = 4(0)-(0^3/3)+C1
=>
C1=2mm-t^-1
x(t)=∫x'(t)dt
=∫(4t-(t^3/3)+2) dt
=2t²-(1/12)t^4+2t+C2
x(0)=0 => C2=0
=>
x(t)=2t² -(1/12)t^4 +2t
(in millimetres)
To find the exact equation for the position of the particle, we need to integrate the given equation for acceleration.
Given:
Acceleration, a = 4 - t^2
Initial velocity, v₀ = 2 mm/s
Initial position, x₀ = 0
Time, t
First, we integrate the acceleration function to find the velocity function:
v = ∫(4 - t^2) dt
v = 4t - (t^3)/3 + C₁
Using the initial velocity, v₀ = 2 mm/s, we can solve for the constant C₁:
2 = 4(0) - (0)^3/3 + C₁
2 = 0 + C₁
C₁ = 2
Thus, the velocity function becomes:
v = 4t - (t^3)/3 + 2
Now, we need to integrate the velocity function to find the position function:
x = ∫v dt
x = ∫(4t - t^3/3 + 2) dt
x = 2t^2 - (t^4)/12 + 2t + C₂
Using the initial position, x₀ = 0, we can solve for the constant C₂:
0 = 2(0)^2 - (0^4)/12 + 2(0) + C₂
0 = 0 + 0 + 0 + C₂
C₂ = 0
Therefore, the exact equation for the position of the particle is:
x = 2t^2 - (t^4)/12 + 2t
Note: The units for time and position are not specified in the question, so the position equation is written in terms of t only. Make sure to use consistent units when solving problems involving physical quantities.