if a particle moves along a horizontal path with an acceleration of a=(2t-9) m/s2 ( t is in seconds), determine the followings.
A) velocity and position of particle at t=9s
B) total distnace travelled by particle
C) average velocity and average speed of the particle
at t=0 s=5m and Vo= 9m/s
a = dv/dt = 2t-9
v = ds/dt = t^2 - 9t + v0
s = 1/3 t^3 - 9/2 t^2 + v0t + s0
fill these in, and the other questions should be easy. Come back with your work if you get stuck.
v = initial v + integral a dt
x = initial x + integral v dt
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v = 9 + integral a dt = t^2 - 9 t + 9
x = 5 + t^3/3 - (9/2)t^2 + 9 t
To solve the given problem, we need to integrate the acceleration function to find the velocity function and position function of the particle. Then we can use those functions to find the required values.
A) Velocity and position of the particle at t = 9s:
1. Integration of acceleration function:
a = 2t - 9
∫a dt = ∫(2t - 9) dt
v = t^2 - 9t + C1
2. Use initial condition V₀ = 9 m/s at t = 0s:
9 = 0^2 - 9(0) + C1
C1 = 9
3. Substitute t = 9 into the velocity function:
V = (9)^2 - 9(9) + 9
V = 81 - 81 + 9
V = 9 m/s
4. Integration of velocity function:
∫v dt = ∫(t^2 - 9t + 9) dt
s = (1/3)t^3 - (9/2)t^2 + 9t + C2
5. Use initial condition s₀ = 5m at t = 0s:
5 = (1/3)(0^3) - (9/2)(0^2) + 9(0) + C2
C2 = 5
6. Substitute t = 9 into the position function:
s = (1/3)(9^3) - (9/2)(9^2) + 9(9) + 5
s = 243 - 365.5 + 81 + 5
s = -36.5 m
Therefore, at t = 9s, the velocity of the particle is 9 m/s and the position of the particle is -36.5 m.
B) Total distance traveled by the particle:
To find the total distance traveled, we need to consider both positive and negative displacement.
1. Displacement from t = 0s to t = 9s:
s₁ = ∫[(1/3)t^3 - (9/2)t^2 + 9t + 5] dt (from t = 0 to t = 9)
s₁ = [(1/12)t^4 - (3/4)t^3 + (9/2)t^2 + 5t] (from 0 to 9)
s₁ = [(1/12)(9^4) - (3/4)(9^3) + (9/2)(9^2) + 5(9)] - [(1/12)(0^4) - (3/4)(0^3) + (9/2)(0^2) + 5(0)]
s₁ = 729/4 - 243/4 + 243/2 + 45
s₁ = 486/4 + 243/2 + 45
s₁ = 121.5 + 121.5 + 45
s₁ = 288 meters
2. Displacement from t = 9s to t = 0s (negative displacement):
s₂ = ∫[(1/3)t^3 - (9/2)t^2 + 9t + 5] dt (from t = 9 to t = 0)
s₂ = [(1/12)t^4 - (3/4)t^3 + (9/2)t^2 + 5t] (from 9 to 0)
s₂ = 0 - [(1/12)(9^4) - (3/4)(9^3) + (9/2)(9^2) + 5(9)]
s₂ = 0 - (729/4 - 243/4 + 243/2 + 45)
s₂ = -486/4 - 243/2 - 45
s₂ = -121.5 - 121.5 - 45
s₂ = -288 meters
Total distance traveled = |s₁| + |s₂| = 288 + 288 = 576 meters
The total distance traveled by the particle is 576 meters.
C) Average velocity and average speed of the particle:
To find the average velocity and average speed, we need to consider the total displacement and the total time interval.
1. Total displacement = s₁ + s₂ = 288 - 288 = 0 meter
2. Total time interval = t = 9s - 0s = 9s
Average velocity = Total displacement / Total time interval
Average velocity = 0 / 9 = 0 m/s
Average speed = Total distance traveled / Total time interval
Average speed = 576 / 9 = 64 m/s
Therefore, at t = 9s, the average velocity of the particle is 0 m/s and the average speed of the particle is 64 m/s.
To find the velocity and position of the particle at t = 9s, we need to integrate the given acceleration function and apply the initial conditions.
A) Velocity at t = 9s:
To find the velocity, we need to integrate the acceleration function with respect to time:
v = ∫(2t - 9) dt
Integrating, we get:
v = t^2 - 9t + C
Now, apply the initial condition at t = 0 and v = 9m/s:
9 = 0^2 - 9(0) + C
C = 9
So, the velocity function becomes:
v = t^2 - 9t + 9
Now substitute t = 9 to get the velocity at t = 9s:
v = 9^2 - 9(9) + 9
v = 81 - 81 + 9
v = 9 m/s
B) Position at t = 9s:
To find the position, we need to integrate the velocity function with respect to time:
s = ∫(t^2 - 9t + 9) dt
Integrating, we get:
s = (t^3/3) - (9t^2/2) + 9t + C
Now, apply the initial condition at t = 0 and s = 5m:
5 = (0^3/3) - (9(0)^2/2) + 9(0) + C
C = 5
So, the position function becomes:
s = (t^3/3) - (9t^2/2) + 9t + 5
Now substitute t = 9 to get the position at t = 9s:
s = (9^3/3) - (9(9)^2/2) + 9(9) + 5
s = 243 - 729/2 + 81 + 5
s = -49.5 m
Hence, the velocity of the particle at t = 9s is 9 m/s and the position is -49.5 m.
C) Average velocity and average speed:
To find the average velocity, we need to find the displacement and divide it by the time elapsed:
Average velocity = (s_final - s_initial) / (t_final - t_initial)
Using the given initial conditions and substituting t_final = 9s, s_final = -49.5m, t_initial = 0s, s_initial = 5m:
Average velocity = (-49.5 - 5) / (9 - 0) = -54.5 / 9 = -6.06 m/s
To find the average speed, we need to find the total distance traveled and divide it by the time elapsed:
Average speed = Total distance / (t_final - t_initial)
Total distance traveled can be obtained by integrating the absolute value of velocity with respect to time from t = 0 to t = 9s:
Total distance = ∫|(2t - 9)| dt from 0 to 9
Integrating, we get:
Total distance = ∫(2t - 9) dt from 0 to 9
Total distance = |t^2 - 9t| from 0 to 9 = |81 - 81| = 0 m
Average speed = 0 / 9 = 0 m/s
Hence, the average velocity of the particle is -6.06 m/s and the average speed is 0 m/s.