A student takes some data with a cart rolling on an inclined track and produces the following set of data: From the graphs and line fit, determine the following quantities: The position of the cart at time t = 0: x0 = 0.3 m Correct: Your answer is correct. The initial velocity: v0 = 0.30 m/s Correct: Your answer is correct. The acceleration: a = 0.205 m/s2 Correct: Your answer is correct. The position at time t = 0.40 s: x(0.40 s)= 0.904 m Incorrect: Your answer is incorrect. The velocity at time t = 0.50 s: v(0.50 s)= 0.403 m/s Correct: Your answer is correct.

Respuesta :

Answer:

[tex]x_f = 0.4364 m[/tex]

Explanation:

As we know that initial velocity of the cart is given as

[tex]v_i = 0.30 m/s[/tex]

initial position is given as

[tex]x_i = 0.30 m[/tex]

acceleration of the cart is given as

[tex]a = 0.205 m/s^2[/tex]

now position of the cart after t= 0.40 s is given as

[tex]x_f = x_i + v_i t + \frac{1}{2}at^2[/tex]

[tex]x_f = 0.30 + 0.30(0.40) + \frac{1}{2}(0.205)(0.40^2)[/tex]

[tex]x_f = 0.4364 m[/tex]

ACCESS MORE