A toy car accelerated from rest for 2.8 seconds, then continued at its maximum speed of 1.8 m s-1 until 8.2 seconds, then put on its brakes and decelerated to a stop in 12.5 seconds. Its motion is plotted on the graph below. How far did it travel when it accelerated ([math: t1] t_1 )? Give your answer to two decimal places.

Question

BlackTom AI · Accepted Answer

First, restate what is being asked: find the distance traveled during the acceleration phase t1, when the toy car starts from rest and reaches a maximum speed of 1.8 m/s in t1 = 2.8 s.

Step 1: Determine the acceleration during the speeding-up phase. Since final speed v = 1.8 m/s is reached from rest (u = 0) in t = 2.8 s, the average acceleration a is a = (v - u) / t = 1.8 / 2.8 ≈ 0.6429 m/s^2.

Step 2: Use the kinematics formula for distance under constant acceleration from rest: s = ut + (1/2) a t^2. With u = 0, this simplifies to s = (1/2) a t^2.

Step 3: Compute s with a ≈ 0.6429 m/s^2 and t = 2.8 s. First calculate t^2: (2.8)^2 = 7.84. Then s = 0.5 × 0.6429 × 7.84.

Step 4: Multiply: 0.6429 × 7.84 ≈ 5.04. Half of that is s ≈ 2.52 m.

Step 5: Round to two decimal places as requested: 2.52 m.

Notes on potential pitfalls: If someone uses an average speed for the acceleration phase, they might incorrectly compute distance as (1/2) v t, which would be (1/2) × 1.8 × 2.8 ≈ 2.52 m as well in this particular case, but this coincidence only works because the motion starts from rest and the acceleration is constant. The robust method is s = (1/2) a t^2 with a = v/t, which yields the same result here. Remember that this distance is only for the acceleration interval; the problem may include other phases with different motion, but they do not affect this specific computation.

MUF0121 Physics Unit 1 - Semester 2, 2025 2.10 Graphical analysis of motion

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