Answer:
56.3 m
Explanation:
aₓ(t) = -0.0320 m/s³ (15.0 s − t)
Integrate to get velocity:
∫ dv = ∫ a dt
vₓ(t) − v₀ₓ = ∫₀ᵗ aₓ(t) dt
vₓ(t) − v₀ₓ = ∫₀ᵗ -0.0320 m/s³ (15.0 s − t) dt
vₓ(t) − v₀ₓ = -0.0320 m/s³ (15.0 s t − ½ t²) |₀ᵗ
vₓ(t) − 8.90 m/s = -0.0320 m/s³ (15.0 s t − ½ t²)
vₓ(t) = -0.0320 m/s³ (15.0 s t − ½ t²) + 8.90 m/s
vₓ(t) = -0.480 m/s² t + 0.0160 m/s³ t² + 8.90 m/s
Integrate again to get position:
∫ dx = ∫ v dt
x(t) − x₀ = ∫₀ᵗ vₓ(t) dt
x(t) − x₀ = ∫₀ᵗ (-0.480 m/s² t + 0.0160 m/s³ t² + 8.90 m/s) dt
x(t) − x₀ = (-0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t) |₀ᵗ
x(t) − (-14.0 m) = -0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t
x(t) = -0.240 m/s² t² + 0.00533 m/s³ t³ + 8.90 m/s t − 14.0 m
Evaluate at t = 10 s:
x(10) = -0.240 m/s² (10 s)² + 0.00533 m/s³ (10 s)³ + 8.90 m/s (10 s) − 14.0 m
x(10) = -24.0 m + 5.33 m + 89.0 m − 14.0 m
x(10) = 56.3 m
As given in the question, the position, velocity and acceleration of the small object are all functions of time, and they all depend on time.
The position (i.e. x coordinate) of the small object at 10.0 seconds is 56.3 meters
Given that:
[tex]a_x(t) = -(0.0320m/s^3)(15.0s - t)[/tex]
Velocity is the integral of acceleration.
So, we have:
[tex]v_x(t) = \int\limits a_x(t)[/tex]
This gives:
[tex]v_x(t) = \int\limits -(0.0320m/s^3)(15.0s - t)[/tex]
Bring out constant
[tex]v_x(t) = -\int\limits (0.0320m/s^3)(15.0s - t)[/tex]
Open bracket
[tex]v_x(t) = -\int\limits 0.48m/s^2 - 0.0320tm/s^3[/tex]
Integrate with respect to t
[tex]v_x(t) = -[0.48t m/s^2 - 0.0320 \frac{t^2}{2}m/s^3 ] + v_x(0)[/tex]
From the question, we understand that:
[tex]v_0x = 8.90m/s[/tex] --- this represents the velocity at [tex]t = 0[/tex]
The equation becomes
[tex]v_x(t) = -[0.48t m/s^2 -0.016t^2m/s^3 ] + 8.90m/s[/tex]
Open bracket
[tex]v_x(t) = -0.48t m/s^2 + 0.016t^2m/s^3 + 8.90m/s[/tex]
Distance (i.e. position) is the integral of velocity.
So, we have:
[tex]x(t) = \int\limits v_x(t)[/tex]
This gives:
[tex]x(t) = \int\limits -0.48t m/s^2 + 0.016t^2m/s^3 + 8.90m/s[/tex]
Integrate with respect to t
[tex]x(t) = \frac{-0.48t^2}{2} m/s^2 + \frac{0.016t^3}{3}m/s^3 + 8.90tm/s + x(0)[/tex]
From the question, we understand that:
[tex]x(0) = -14.0m[/tex] --- this represents the initial position
So, we have:
[tex]x(t) = \frac{-0.48t^2}{2} m/s^2 + \frac{0.016t^3}{3}m/s^3 + 8.90tm/s -14.0m[/tex]
Substitute 10.0s for t to calculate the object position
[tex]x(10.0s) = \frac{-0.48 \times (10.0s)^2}{2} m/s^2 + \frac{0.016\times (10.0s)^3}{3}m/s^3 + 8.90\times (10.0s) m/s -14.0m[/tex]
[tex]x(10.0s) = \frac{-0.48 \times (10.0)^2}{2} m + \frac{0.016\times (10.0)^3}{3}m + 8.90\times (10.0) m -14.0m[/tex]
[tex]x(10.0s) = -24.0m + 5.3m + 89.0 m -14.0m[/tex]
[tex]x(10.0s) = 56.3m[/tex]
Hence, the object's position at 10.0 seconds is 56.3 meters
Read more about acceleration, velocity and time at:
https://brainly.com/question/12134554
