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?

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:

Velocity is the integral of acceleration.

So, we have:

This gives:

Bring out constant

Open bracket

Integrate with respect to t

From the question, we understand that:

--- this represents the velocity at

The equation becomes

Open bracket

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

So, we have:

This gives:

Integrate with respect to t

From the question, we understand that:

--- this represents the initial position

So, we have:

Substitute 10.0s for t to calculate the object position

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

Read more about acceleration, velocity and time at:

link

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