Question

quiz on graphs of kinematics
physics

1. a ball is placed at rest at the top of a hill. it travels with constant acceleration for the first 12 seconds and reaches a speed of 4 m/s. it then decelerates at a constant rate of 0.1m/s² for 20 seconds. it then travels at a constant speed for a further 18 seconds.

draw a speed-time graph for the ball over the course of this 50 second.
grid image (velocity vs time axes with grid)

1. below is a speed-time graph of a track cyclist during a race. work out the total distance travelled by the cyclist over the course of the race.

speed-time graph (velocity vs time) with axes and plotted segments: from (0,0) to (10,15), (10,15) to (15,10), (15,10) to (20,20), (20,20) to (50,20)

Answer:

Question 1: Drawing the Speed - Time Graph

Step 1: Analyze the first phase (0 - 12 seconds)

The ball starts from rest (\(u = 0\ m/s\)) and accelerates with constant acceleration until \(t = 12\ s\) when \(v=4\ m/s\). On a speed - time graph, this is a straight line from \((0,0)\) to \((12,4)\) since the acceleration is constant (so the speed - time graph has a constant slope in this interval).

Step 2: Analyze the second phase (12 - 32 seconds)

The ball decelerates at a rate of \(a=- 0.1\ m/s^{2}\). We can find the time taken to stop or the final speed. Using the formula \(v = u+at\), where \(u = 4\ m/s\), \(a=-0.1\ m/s^{2}\). Let's find the time when the speed becomes \(v = 0\) (but in our case, the time for deceleration is 20 seconds). The speed at \(t=12 + 20=32\ s\) is \(v=4+( - 0.1)\times20=4 - 2=2\ m/s\). So the graph from \(t = 12\ s\) to \(t = 32\ s\) is a straight line with a negative slope (deceleration) from \((12,4)\) to \((32,2)\).

Step 3: Analyze the third phase (32 - 50 seconds)

The ball travels at a constant speed. The time duration is \(18\) seconds, so from \(t = 32\ s\) to \(t=32 + 18=50\ s\), the speed remains \(2\ m/s\). So the graph is a horizontal line from \((32,2)\) to \((50,2)\).

Question 2: Calculating the Total Distance Travelled by the Cyclist

The total distance travelled by an object from a speed - time graph is the area under the speed - time curve. We can divide the graph into geometric shapes (triangles, trapezoids, rectangles) and calculate the area of each shape and sum them up.

Step 1: Analyze the first part (0 - 10 seconds)

This is a triangle with base \(b = 10\ s\) and height \(h = 15\ m/s\). The area of a triangle is \(A_1=\frac{1}{2}\times b\times h\). So \(A_1=\frac{1}{2}\times10\times15 = 75\ m\).

Step 2: Analyze the second part (10 - 15 seconds approximately, but looking at the graph, from 10 to the time when speed is 10 m/s. Let's assume the time between 10 and the trough (speed = 10 m/s) is \(t_2\). From the graph, the slope from 10 to the trough: the speed goes from 15 to 10 m/s. Let's assume the time taken is \(t_2 = 5\ s\) (since the graph seems symmetric? Wait, actually, looking at the graph, from \(t = 10\) to \(t\) where speed is 10 m/s, let's say the time is \(t = 15\ s\) (so the base is \(15 - 10=5\ s\), the two parallel sides of the trapezoid are \(15\ m/s\) and \(10\ m/s\)). The area of a trapezoid is \(A_2=\frac{1}{2}\times(b_1 + b_2)\times h\), where \(b_1 = 15\), \(b_2 = 10\), \(h=5\). So \(A_2=\frac{1}{2}\times(15 + 10)\times5=\frac{1}{2}\times25\times5 = 62.5\ m\).

Step 3: Analyze the third part (15 - 20 seconds)

This is a triangle (or a trapezoid) where the speed goes from 10 to 20 m/s. The time duration is \(20 - 15 = 5\ s\). The area of a triangle (since it's a straight line from (15,10) to (20,20)) is \(A_3=\frac{1}{2}\times(20 - 15)\times(20 - 10)+\frac{1}{2}\times(20 - 15)\times10\)? Wait, no. The area of the trapezoid with \(b_1 = 10\), \(b_2 = 20\), \(h = 5\) is \(A_3=\frac{1}{2}\times(10 + 20)\times5=\frac{1}{2}\times30\times5 = 75\ m\).

Step 4: Analyze the fourth part (20 - 50 seconds)

This is a rectangle with length \(l=50 - 20 = 30\ s\) and width \(w = 20\ m/s\). The area of a rectangle is \(A_4=l\times w=30\times20 = 600\ m\).

Step 5: Sum up the areas

Total distance \(d=A_1+A_2+A_3+A_4\)
\(d = 75+62.5 + 75+600\)
\(d=812.5\ m\) (Wait, maybe a better way: Let's re - analyze the graph:

Alternative approach:

• From \(t = 0\) to \(t = 10\): Triangle with base 10, height 15. Area \(=\frac{1}{2}\times10\times15 = 75\)
• From \(t = 10\) to \(t = 15\): Triangle with base 5, height \(15 - 10 = 5\) (but it's a decreasing slope). Area \(=\frac{1}{2}\times5\times5 = 12.5\)? No, wait the speed goes from 15 to 10, so the area is the area of the trapezoid with bases 15 and 10, height 5. \(A=\frac{(15 + 10)}{2}\times5=62.5\) (same as before)
• From \(t = 15\) to \(t = 20\): Triangle with base 5, height \(20 - 10 = 10\). Area \(=\frac{1}{2}\times5\times10 = 25\)? No, the speed goes from 10 to 20, so trapezoid with bases 10 and 20, height 5. \(A=\frac{(10 + 20)}{2}\times5 = 75\)
• From \(t = 20\) to \(t = 50\): Rectangle with length \(50 - 20=30\), width 20. Area \(=30\times20 = 600\)

Total distance \(=75+62.5 + 75+600=812.5\ m\)

Final Answers

Question 1

The speed - time graph has three segments:

• From \((0,0)\) to \((12,4)\) (constant acceleration)
• From \((12,4)\) to \((32,2)\) (constant deceleration)
• From \((32,2)\) to \((50,2)\) (constant speed)

Question 2

The total distance travelled by the cyclist is \(\boldsymbol{812.5\ m}\) (or depending on the exact graph interpretation, if we consider the time intervals more accurately, but based on the given graph, this is the calculation).