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 ($ t_1 $)? Give your answer to two decimal places.

Question

BlackTom AI · Accepted Answer

First, identify the phase of motion for t = t1 where the car is accelerating from rest. The car starts from rest and reaches a maximum speed of 1.8 m/s in 2.8 s, so the acceleration is constant during this interval: a = Δv/Δt = 1.8 / 2.8 ≈ 0.642857 m/s^2.
Next, compute the distance traveled during this acceleration phase using the kinematics formula for uniformly accelerated motion starting from rest: s = (1/2) a t^2.
Plugging in the values: s = 0.5 × 0.642857 × (2.8)^2. Since (2.8)^2 = 7.84 and 0.5 × 0.642857 ≈ 0.3214285, the product is s ≈ 0.3214285 × 7.84 ≈ 2.519999 ≈ 2.52 m when rounded to two decimals.
Therefore, the distance covered during the acceleration period t1 is about 2.52 m. 
Note: If one checks consistency with the graph, the triangular/linear rise corresponds to this initial acceleration region, and the plateau thereafter corresponds to the constant speed before deceleration.

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