The position-time graph for a bug crawling along a line is shown in item 4 below. Determine whether the velocity is positive, ne
gative, or zero at each of the times marked on the graph.
1 answer:
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
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height = 45 m
Explanation:
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Final velocity = 30 m/s
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Using the eqn of motion
where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height
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Answer:
Yes, the car driver is exceeding the given limit.
Explanation:
<u>Given:</u>
- Speed of the car, v = 38.0 m/s.
- Speed limit of the highway,
We know,
1 mi = 1.60934 km.
1 km = 1000 m.
Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.
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1 minute = 60 seconds.
Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.
Using these values,
Therefore,
Clearly,
which means, the car driver is exceeding the given speed limit.