A particle moves on the x-axis with an acceleration, a=(6t-4)ms⁻1. Find the position and velocity of the particle at t=3 , if the particle is at origin and has a velocity of when t=0
This is an IVP (initial value problem).
x"(t)=(6t-4) ms-2 ....(1)
"...if the particle is at origin and has a velocity of ? when t=0."
Integrate (1) with respect to t to find x'(t) and x(t). Do not forget the integration constants at each integration.
Use the initial conditions
x(0)=0 and x'(0)=(???)
to find C1 and C2 (int. constants).
When C1 and C2 are known, evaluate
x(3) and x'(3) as required.
thanzz again
To find the position and velocity of the particle at t = 3, we'll need to integrate the given acceleration function with respect to time.
Given:
Acceleration, a = (6t - 4) m/s²
First, we need to find the velocity of the particle by integrating the acceleration function with respect to time. The velocity function, v(t), can be found by integrating the acceleration function, a(t):
v(t) = ∫ a(t) dt
Integrating the given acceleration, a = (6t - 4) m/s², with respect to time, we get:
v(t) = ∫ (6t - 4) dt
= 3t² - 4t + C
where C is the constant of integration. To determine the value of C, we use the initial condition given that the particle has a velocity of 0 (v(0) = 0) when t = 0:
v(0) = 3(0)² - 4(0) + C
= 0 + 0 + C
= C
Therefore, C = 0.
Now we have the velocity function:
v(t) = 3t² - 4t
To find the position function, x(t), we integrate the velocity function with respect to time:
x(t) = ∫ v(t) dt
Integrating the velocity function, v(t) = 3t² - 4t, with respect to time, we get:
x(t) = ∫ (3t² - 4t) dt
= t³ - 2t² + C'
where C' is the constant of integration. To determine the value of C', we use the initial condition given that the particle is at the origin (x(0) = 0) when t = 0:
x(0) = (0)³ - 2(0)² + C'
= 0 - 0 + C'
= C'
Therefore, C' = 0.
Now we have the position function:
x(t) = t³ - 2t²
To find the position and velocity of the particle at t = 3, we substitute t = 3 into the position and velocity functions:
Position at t = 3:
x(3) = (3)³ - 2(3)²
= 27 - 18
= 9 meters
Velocity at t = 3:
v(3) = 3(3)² - 4(3)
= 27 - 12
= 15 m/s
Therefore, the position of the particle at t = 3 is 9 meters, and its velocity at t = 3 is 15 m/s.