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a public transportation bus travels back and forth along a road, following the position versus time graph
over the time period shown in the graph. there are five points labeled on the graph.
point 1(0, 40), point 2(1,15), point 3(1.5, 30), p4(3, 35), p5(5, 60).
(a) what is the average speed and velocity (magnitude and direction) of the bus between point 1 and point 3?
(b) what is the average speed and velocity (magnitude and direction) of the bus between point 3 and point 5?
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In each pair, first number is the time t (horizontal axis);
second number is the coordinate of the position y (vertical axis).
y = y(t) is the given graph.
The time is given in abstract no-named units of time.
The position is also given in abstract no-named units of the length.
For question (a), the bus was at y= 40 at the time t= 0;
the bus was at y= 30 at the time t= 1.5.
We know that average speed is the traveled distance divided by the travel time.
So, the traveled distance in question (a) is 30 - 40 = -10 units (10 units in negative direction);
the travel time is 1.5 - 0 = 1.5 units.
Thus the magnitude of the average speed is = 6.667 units of speed (rounded),
and the direction of velocity is opposite to the axis of coordinate (y-axis on the graph).
For question (b), the bus was at y= 30 at the time t= 1.5;
the bus was at y= 60 at the time t= 5.
Again, we know that average speed is the traveled distance divided by the travel time.
So, the traveled distance in question (a) is 60 - 30 = 30 units;
the travel time is 5 - 1.5 = 3.5 units.
Thus the magnitude of the average speed is = 8.571 units of speed (rounded),
and the direction of velocity is along the axis of coordinate (y-axis on the graph).
Solved.
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