Physics, 23.09.2019 17:30 Shamasen1

Asmall object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). at t = 0 the object is at x = -14.0 m and has velocity v0x = 8.90 m/s. what is the x-coordinate of the object when t = 10.0 s?

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Answer from: kidyugey

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:

a_x(t) = -(0.0320m/s^3)(15.0s - t)

Velocity is the integral of acceleration.

So, we have:

$v_x(t) = \int\limits a_x(t)$

This gives:

$v_x(t) = \int\limits -(0.0320m/s^3)(15.0s - t)$

Bring out constant

$v_x(t) = -\int\limits (0.0320m/s^3)(15.0s - t)$

Open bracket

$v_x(t) = -\int\limits 0.48m/s^2 - 0.0320tm/s^3$

Integrate with respect to t

$v_x(t) = -[0.48t m/s^2 - 0.0320 \frac{t^2}{2}m/s^3 ] + v_x(0)$

From the question, we understand that:

v_0x = 8.90m/s --- this represents the velocity at t = 0

The equation becomes

v_x(t) = -[0.48t m/s^2 -0.016t^2m/s^3 ] + 8.90m/s

Open bracket

v_x(t) = -0.48t m/s^2 + 0.016t^2m/s^3 + 8.90m/s

Distance (i.e. position) is the integral of velocity.

So, we have:

$x(t) = \int\limits v_x(t)$

This gives:

$x(t) = \int\limits -0.48t m/s^2 + 0.016t^2m/s^3 + 8.90m/s$

Integrate with respect to t

$x(t) = \frac{-0.48t^2}{2} m/s^2 + \frac{0.016t^3}{3}m/s^3 + 8.90tm/s + x(0)$

From the question, we understand that:

x(0) = -14.0m --- this represents the initial position

So, we have:

$x(t) = \frac{-0.48t^2}{2} m/s^2 + \frac{0.016t^3}{3}m/s^3 + 8.90tm/s -14.0m$

Substitute 10.0s for t to calculate the object position

$x(10.0s) = \frac{-0.48 \times (10.0s)^2}{2} m/s^2 + \frac{0.016\times (10.0s)^3}{3}m/s^3 + 8.90\times (10.0s) m/s -14.0m$

$x(10.0s) = \frac{-0.48 \times (10.0)^2}{2} m + \frac{0.016\times (10.0)^3}{3}m + 8.90\times (10.0) m -14.0m$

x(10.0s) = -24.0m + 5.3m + 89.0 m -14.0m

x(10.0s) = 56.3m

Hence, the object's position at 10.0 seconds is 56.3 meters

Read more about acceleration, velocity and time at:

link

Answer from: jessiebotello7209

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

Answer from: Quest

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Answer from: Quest

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