Final answer:
Acceleration calculation requires further clarification on timing interpretation; the naive approach suggests a deceleration, which contradicts the expected result from the scenario described.
Step-by-step explanation:
To determine the acceleration of the cart, we'll use the photo-gate timer readings to calculate the average velocities for each segment and then use these to calculate the acceleration. The cart's length is mentioned as 12 cm (which we convert to 0.12 m) and the distance between the two gates is 0.70 m.
The average velocity as the cart passes each gate can be computed using the formula:
V = d / t
For the first gate (Gate 1), the cart travels a distance equal to the length of the cart plus the distance between the gates, while for the second gate (Gate 2), the distance is just the length of the cart:
- V1 = (0.70 + 0.12) / 0.30 s = 2.733 m/s
- V2 = 0.12 m / 0.20 s = 0.60 m/s
Now, we will use the average velocities to find the acceleration (a) of the cart:
a = (V2 - V1) / t
Since we are not given the time interval between the two readings, we will assume the time to be the interval between the two times recorded (0.30 s - 0.20 s = 0.10 s), yielding:
a = (0.60 m/s - 2.733 m/s) / 0.10 s = -21.33 m/s2
The negative sign indicates the cart is decelerating. However, we expected a positive acceleration since the cart is being pulled by a mass falling due to gravity. The discrepancy implies that our assumption about taking the time difference as 0.10 s may not be correct. Instead, the time of 0.30 s and 0.20 s could refer to the time the front of the cart enters and leaves the gate, respectively. Since the question requires the acceleration, an additional clarification on the interpretation of the timing might be necessary to provide an accurate answer.