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AQA GCSE · Question 05.3 · Forces

The student held the trolley stationary and then released it. The trolley moved along the runway with a constant acceleration. The student recorded the time taken for the trolley to travel a measured distance along the runway. Describe how the acceleration of the trolley can be calculated using the time taken and distance travelled by the trolley.

How to approach this question

1. Think of the equations of motion (SUVAT equations). Which equation links distance (s), initial velocity (u), time (t), and acceleration (a)? 2. What is the initial velocity of the trolley if it is released from rest? 3. How can you simplify the equation from step 1 using this information? 4. Rearrange the simplified equation to solve for acceleration (a).

Full Answer

1. The student should use the equation of motion: s = ut + ½at². 2. Since the trolley starts from rest, its initial velocity (u) is 0. The equation simplifies to s = ½at². 3. The student can rearrange this equation to make acceleration (a) the subject: a = 2s / t². 4. By substituting their measured values for distance (s) and time (t), they can calculate the acceleration.
To find the acceleration from distance and time, we can use one of the standard equations of motion. The relevant equation is: s = ut + ½at² where: s = distance travelled u = initial velocity t = time taken a = acceleration Since the trolley is released from a stationary position, its initial velocity (u) is 0 m/s. This simplifies the equation because the `ut` term becomes zero: s = ½at² To calculate the acceleration, we need to rearrange this equation to make `a` the subject: 2s = at² a = 2s / t² The student measures the distance `s` and the time `t`, then uses this final equation to calculate the acceleration `a`.

Common mistakes

✗ Using the wrong equation, like v = u + at, which would require knowing the final velocity.\n✗ Forgetting that the initial velocity is zero.\n✗ Rearranging the formula incorrectly.

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