Question

Determine the horizontal position x of the cart at t=9.0 s if the cart is located at x =2 m at t = 0

Answer 1

Answer:

1.1m

Explanation:

Given values: time t = 9s, initial position x₀ = 2m, Values from the graph:

initial velocity v₀= 0.8m/s, constant acceleration a = -0.2m/s²(slope of the graph between time t = 0 and t = 9s)

The equation for position x and a constant acceleration is given by:

$x=\frac{1}{2}at^2+v_0t+x_0$

Using the given values to solve for x:

$x=\frac{1}{2}\times(-\frac{2}{10})\times9^2+\frac{8}{10}\times9 +2\\ \\x=-\frac{81}{10} +\frac{72}{10} +\frac{20}{10}=\frac{11}{10}$

Answer 2

Horizontal position X is 1.1 m

Given:

Initial time t = 0 sec

Final time T = 9 sec

Initial velocity u = 0.8 m/s

Final velocity v = -1 m/s

Initial position X₀ = 2 m

Find:

Horizontal position X

Computation:

We know that;

a = (v - u) / Δt

So,

a = (-1 - 0.8) / (9 - 0)

a = -1.8 / 9

a = -0.2 m/s²

The equation for position X

X = ut + (1/2)(a)(t²) + X₀

X = (0.8)(9) + (1/2)(-0.2)(9²) + 2

X = 7.2 + (-0.1)(81) + 2

X = 9.2 - 8.1

X = 1.1 m

Horizontal position X = 1.1 m

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