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 calculate the distance travelled by the cyclist between points Y and Z, we first determine the areas of the geometrical shapes involved. The rectangle has an area of 108 m, and the triangle has an area of 54 m. The total area, which represents the distance, is obtained by adding these two values:

Total Area = Area of Rectangle + Area of Triangle = 108 m + 54 m = 162 m. Thus, the distance travelled by the cyclist between Y and Z is 162 m.

The force applied on the pedal generates a torque about the pedal axle, which is the pivot point. As the pedal rotates, this force generates a downward force on the chain, which in turn creates a moment about the rear axle. This moment can be described by the formula:

$ext{Moment} = ext{Force} imes ext{Distance}$ where the distance is the horizontal distance from the pedal axle to the rear axle.

To compute the initial acceleration of the cyclist, we can use the formula:

$a = \frac{(v_f - v_i)}{t}$ where:

• $v_f$ = final velocity = 2.4 m/s
• $v_i$ = initial velocity = 0 m/s
• $t$ = time taken to accelerate.

Using the distance travelled during acceleration, we know:

$d = \frac{(v_i + v_f)}{2} \times t$ Substituting the known values:

$0.018 km = \frac{(0 + 2.4)}{2} \times t$ Converting 0.018 km to meters gives 18 m. Therefore:

$18 = 1.2 \times t$ This means: $t = \frac{18}{1.2} = 15 s.$

Substituting $t$ back into the acceleration formula:

$a = \frac{(2.4 - 0)}{15} = 0.16 m/s^2.$

To find the resultant force, we draw a vector diagram using the horizontal and vertical components. The horizontal force is 200 N and the vertical force is 75 N. The resultant force can be calculated using Pythagoras' theorem:

$F_r = \sqrt{(200^2 + 75^2)} = \sqrt{40000 + 5625} = \sqrt{45625} = 213.6 N.$

The direction can be calculated using the tangent function:

$\theta = \tan^{-1}\left(\frac{75}{200}\right) = 20.22^{\circ}.$

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