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mars1129 [50]
3 years ago
11

A person takes a trip, driving with a constant

Physics
1 answer:
podryga [215]3 years ago
8 0
For this problem, we are asked to calculate for the distance traveled. We set up the equations as follows:
Distance = 61 km/hr * (time + 20.8/60 s) Distance = 98.5 km/hr * time

We equate the two equations, then we can solve for the time spent on the trip. Hope this answers the question. Have a nice day.
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A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The
mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

v = v_{o}+a\cdot t

Where:

v_{o} - Initial velocity, measured in meters per second.

v - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t - Time, measured in seconds.

Now, we clear acceleration within expression:

a = \frac{v-v_{o}}{t}

Initial and final velocities are now converted from kilometers per hour into meters per second:

v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v_{o} = 30\,\frac{m}{s}

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v = 37.5\,\frac{m}{s}

If we know that t = 3.5\,s, then, the average acceleration of the car is:

a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}

a = 2.143\,\frac{m}{s^{2}}

The average acceleration of the car is 2.143 meters per square second.

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Do you have multiple choice i can see?

Explanation:

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