A flat go-kart track consists of two straightways 100 m long with semicircular ends which have a radius of 25.0 m. A go-kart executes the curved end with a constant tangential acceleration while SLOWING from 12.5 m/s to 5.0 m/s. Determine the magnitude of the total acceleration of the go-kart at the beginning and end of the curve.
To determine the magnitude of the total acceleration at the beginning and end of the curve, you need to consider the constant tangential acceleration and use the equations of motion.
First, let's find the centripetal acceleration at the beginning and end of the curve. The centripetal acceleration is given by the equation:
a_c = v^2 / r
where a_c is the centripetal acceleration, v is the velocity, and r is the radius.
At the beginning of the curve:
v = 12.5 m/s
r = 25.0 m
Substituting the values into the equation, we get:
a_c_beginning = (12.5 m/s)^2 / 25.0 m
a_c_beginning = 6.25 m^2/s^2 / 25.0 m
a_c_beginning = 0.25 m/s^2
Similarly, at the end of the curve:
v = 5.0 m/s
r = 25.0 m
Substituting the values into the equation, we get:
a_c_end = (5.0 m/s)^2 / 25.0 m
a_c_end = 1.0 m^2/s^2 / 25.0 m
a_c_end = 0.04 m/s^2
Now let's find the tangential acceleration. The change in velocity is given by:
Δv = v_final - v_initial
At the beginning of the curve:
v_initial = 12.5 m/s
v_final = 5.0 m/s
Substituting the values into the equation, we get:
Δv_beginning = 5.0 m/s - 12.5 m/s
Δv_beginning = -7.5 m/s
Similarly, at the end of the curve:
v_initial = 5.0 m/s
v_final = 12.5 m/s
Substituting the values into the equation, we get:
Δv_end = 12.5 m/s - 5.0 m/s
Δv_end = 7.5 m/s
Now, let's calculate the tangential acceleration using the equation:
a_t = Δv / t
where a_t is the tangential acceleration and t is the time taken to change the velocity.
Since the problem states that the go-kart is slowing down, we can assume that the time taken to change velocity is the same for both the beginning and end of the curve.
For convenience, let's assume that t = 1 second.
At the beginning of the curve:
Δv_beginning = -7.5 m/s
t = 1 s
Substituting the values into the equation, we get:
a_t_beginning = (-7.5 m/s) / (1 s)
a_t_beginning = -7.5 m/s^2
Similarly, at the end of the curve:
Δv_end = 7.5 m/s
t = 1 s
Substituting the values into the equation, we get:
a_t_end = (7.5 m/s) / (1 s)
a_t_end = 7.5 m/s^2
Finally, to find the total acceleration at the beginning and end of the curve, we can use the formula for vector addition:
a_total = √(a_c^2 + a_t^2)
At the beginning of the curve:
a_c_beginning = 0.25 m/s^2
a_t_beginning = -7.5 m/s^2
Substituting the values into the formula, we get:
a_total_beginning = √((0.25 m/s^2)^2 + (-7.5 m/s^2)^2)
a_total_beginning = √(0.0625 m^2/s^4 + 56.25 m^2/s^4)
a_total_beginning = √56.3125 m^2/s^4
a_total_beginning ≈ 7.508 m/s^2
Similarly, at the end of the curve:
a_c_end = 0.04 m/s^2
a_t_end = 7.5 m/s^2
Substituting the values into the formula, we get:
a_total_end = √((0.04 m/s^2)^2 + (7.5 m/s^2)^2)
a_total_end = √(0.0016 m^2/s^4 + 56.25 m^2/s^4)
a_total_end = √56.2516 m^2/s^4
a_total_end ≈ 7.503 m/s^2
Therefore, the magnitude of the total acceleration at the beginning of the curve is approximately 7.508 m/s^2, and at the end of the curve is approximately 7.503 m/s^2.