Question

speed and velocity
directions: draw a line on each of the time-distance graphs below as
1
distance (km)
0 15 30 45 60
time (minutes)
show a car’s constant speed of 75 km/h
on a city street.
answer each question on the lines prov

Explanation:

Step1: Recall the formula for distance

The formula relating speed (\(v\)), time (\(t\)) and distance (\(d\)) is \(d = v\times t\). Here, the speed \(v = 75\space km/h\). We need to convert time from minutes to hours because the speed is in km per hour.

Step2: Convert time units

Let's take a time value, say \(t = 60\) minutes. Since \(60\) minutes \( = 1\) hour.

Step3: Calculate distance for \(t = 60\) minutes

Substitute \(v = 75\space km/h\) and \(t = 1\space h\) into the formula \(d=v\times t\). So \(d = 75\times1=75\space km\). Also, for \(t = 30\) minutes (\(t = 0.5\space h\)), \(d=75\times0.5 = 37.5\space km\), for \(t = 15\) minutes (\(t=0.25\space h\)), \(d = 75\times0.25=18.75\space km\) and for \(t = 45\) minutes (\(t = 0.75\space h\)), \(d=75\times0.75 = 56.25\space km\)

Step4: Plot the points

We have the following points: \((0,0)\) (at time \(0\) minutes, distance is \(0\) km), \((15, 18.75)\), \((30, 37.5)\), \((45, 56.25)\), \((60, 75)\). Now, draw a straight line passing through these points on the given time - distance graph. The line should start at the origin \((0,0)\) and go up to \((60,75)\) (or pass through the other intermediate points) with a constant slope (since speed is constant, the graph of distance vs time for constant speed is a straight line).

Answer:

To show a car's constant speed of \(75\space km/h\) on the time - distance graph:

1. Use the formula \(d=v\times t\) (where \(v = 75\space km/h\) and convert \(t\) from minutes to hours).
2. Calculate distance for different time values:

• At \(t = 0\) min (\(t = 0\space h\)), \(d=0\space km\) (point \((0,0)\)).
• At \(t = 15\) min (\(t=\frac{15}{60}=0.25\space h\)), \(d = 75\times0.25 = 18.75\space km\) (point \((15,18.75)\)).
• At \(t = 30\) min (\(t=\frac{30}{60}=0.5\space h\)), \(d = 75\times0.5=37.5\space km\) (point \((30,37.5)\)).
• At \(t = 45\) min (\(t=\frac{45}{60}=0.75\space h\)), \(d = 75\times0.75 = 56.25\space km\) (point \((45,56.25)\)).
• At \(t = 60\) min (\(t = 1\space h\)), \(d=75\times1 = 75\space km\) (point \((60,75)\)).

1. Draw a straight line passing through the points \((0,0)\), \((15,18.75)\), \((30,37.5)\), \((45,56.25)\) and \((60,75)\) on the given time - distance graph.