Physics 5054/22 — October/November 2016

Walkthrough

During the final $10.0\text{ s}$, the skier travels at a constant speed. According to Newton's first law, an object moving at a constant velocity (constant speed in a straight line) has no acceleration. By Newton's second law, $F = ma$, if acceleration $a = 0$, the resultant force $F$ must also be zero. The driving force (or component of gravity) is exactly balanced by friction and drag.

Key Takeaways

• Constant speed in a straight line implies zero acceleration and therefore a resultant force of zero.
• 'Resultant force' is the vector sum of all forces; it is not the same as 'no forces acting'.

Common Mistakes

• Answering 'no forces' instead of 'zero resultant force'. Forces are still acting (gravity, normal reaction, friction, drag), but they cancel out.
• Confusing 'resultant force' with 'driving force'.

Things to Be Careful About

• The question asks for the 'size' of the resultant force, so a numerical value of '0' or '0 N' is required.

Walkthrough

The motion has two distinct phases:

1. 0 to 5.0 s: Uniform acceleration from rest. The speed increases linearly from $0\text{ m/s}$ at $t = 0\text{ s}$ to $17\text{ m/s}$ at $t = 5.0\text{ s}$. On a speed-time graph, this is a straight line with a positive gradient passing through the origin and the point $(5.0, 17)$.
2. 5.0 s to 15.0 s: Constant speed of $17\text{ m/s}$. The speed does not change, so the graph is a horizontal line at $v = 17\text{ m/s}$ from $t = 5.0\text{ s}$ to $t = 15.0\text{ s}$ (and beyond, up to at least $15.0\text{ s}$).

Key Takeaways

• A straight line with a positive gradient on a speed-time graph represents uniform acceleration.
• A horizontal line on a speed-time graph represents constant speed (zero acceleration).
• The gradient of a speed-time graph is acceleration; the area under it is distance.

Common Mistakes

• Drawing a curve for the first phase instead of a straight line (acceleration is uniform, so the graph must be straight).
• Extending the horizontal line only to $t = 10\text{ s}$; the question asks for the whole $15.0\text{ s}$, so the line must reach at least $t = 15.0\text{ s}$.
• Starting the line at a non-zero speed or not passing through the origin.

Things to Be Careful About

• Read the coordinates accurately: the point is $(5.0, 17)$, not $(5, 15)$ or similar.
• Ensure the graph is drawn with a sharp corner at $(5.0, 17)$, not a smooth curve, as the acceleration changes abruptly from $3.4\text{ m/s}^2$ to $0\text{ m/s}^2$.

Walkthrough

The area under a speed-time graph represents the distance travelled by the object. For this specific graph, the area is a trapezium (or a triangle plus a rectangle), and calculating this area gives the total distance covered during the $15.0\text{ s}$.

Key Takeaways

• The area under a speed-time graph is always equal to the distance travelled.
• This is a fundamental property of integration in kinematics, applicable at O Level as a geometric area calculation.

Common Mistakes

• Saying 'the gradient' gives the distance (the gradient gives acceleration).
• Saying 'the area gives speed' (the value on the y-axis gives speed).

Things to Be Careful About

• The question asks 'how the distance can be determined', so a method is required, not a numerical value. 'Area under the graph' or 'area of the trapezium' are the accepted phrases.