The graph shows the speed at which a remote controlled toy helicopter travels during a test flight along a straight flight path. What is the deceleration (in m s-2) of the helicopter between t2 = 7 seconds and t3 = 11.6 seconds, if it reached a maximum speed of 12.4 m s-1?

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BlackTom AI · Accepted Answer

The problem provides a velocity–time graph for a remote controlled helicopter and asks for the deceleration between t2 = 7 s and t3 = 11.6 s, given that the helicopter reaches a maximum speed of 12.4 m/s.

First, interpret the graph: the helicopter accelerates up to a maximum speed, then maintains that speed for a while, and finally decelerates to a stop. The dashed vertical lines at t2 and t3 delimit the interval during which deceleration occurs. According to the description and the typical trapezoidal shape, the speed drops from the maximum value to zero between t2 and t3.

Next, use the definition of deceleration (which is negative acceleration): a = Δv / Δt.
- Initial velocity at the start of deceleration (at t2): v_i = 12.4 m/s (the maximum speed).
- Final velocity at the end of deceleration (at t3): v_f = 0 m/s (it comes to rest).
- Time interval for deceleration: Δt = t3 − t2 = 11.6 s − 7.0 s = 4.6 s.

Compute the change in velocity: Δv = v_f − v_i = 0 − 12.4 m/s = −12.4 m/s.
Then the deceleration is a = Δv / Δt = (−12.4 m/s) / (4.6 s) ≈ −2.6957 m/s².
Rounding to a reasonable precision, the deceleration is about −2.7 m/s². The negative sign indicates a reduction in speed.

In summary, between t2 = 7 s and t3 = 11.6 s, the helicopter slows from 12.4 m/s to 0 m/s over 4.6 seconds, giving a deceleration of approximately 2.7 m/s² in the opposite direction of motion.

The graph shows the speed at which a remote controlled toy helicopter travels during a test flight along a straight flight path. What is the deceleration (in m s-2) of the helicopter between t2 = 7 seconds and t3 = 11.6 seconds, if it reached a maximum speed of 12.4 m s-1?

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