What name is given to force A? Determine the distance travelled by the cyclist between Y and Z - AQA - GCSE Physics - Question 7 - 2021 - Paper 1

To find the distance travelled by the cyclist between Y and Z, we can use the area under the speed-time graph. The rectangle’s area represents the constant speed from Y to Z.

• Area of rectangle = base × height = (9 m × 12 m) = 108 m.
• Area of triangle = (1/2 × base × height) = (1/2 × 6 m × 9 m) = 27 m.
• Total distance = Area of rectangle + Area of triangle = 108 m + 27 m = 135 m.

The force applied on the pedal creates a torque about the pedal axle. This torque is transmitted through the chain to gear B, generating a force that rotates gear B, which in turn causes a moment about the rear axle of the bicycle. This moment allows the rear wheel to turn, propelling the cyclist forward.

To calculate the initial acceleration of the cyclist, we can use the following kinematic equation:

$a = \frac{v_f - v_i}{t}$ where, v_f = final velocity = 2.4 m/s, v_i = initial velocity = 0 m/s, t = time taken.

From the distance travelled during acceleration:

• Distance = 0.018 km = 18 m
• Using the formula (s = ut + \frac{1}{2}at^2):
• 18 = 0*t + \frac{1}{2} a t^2
• Assuming acceleration is uniform, a = \frac{(v_f^2 - v_i^2)}{2s} = \frac{(2.4^2 - 0^2)}{2 * 18} = 0.08 m/s².

To determine the resultant force, we can use the Pythagorean theorem to find the magnitude:

$\text{Resultant Force} = \sqrt{(F_{horizontal}^2 + F_{vertical}^2)} = \sqrt{(200 N)^2 + (75 N)^2}$

Calculating gives us:

$= \sqrt{40000 + 5625} = \sqrt{45625} \approx 213.6 N$

Direction can be calculated using the tangent function: $\theta = \tan^{-1}\left(\frac{F_{vertical}}{F_{horizontal}}\right) = \tan^{-1}\left(\frac{75}{200}\right) \approx 20.4°$

Thus, the magnitude of the resultant force is approximately 214 N, and the direction is about 20° from the horizontal.