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pshichka [43]
3 years ago
8

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

driver stomps on the gas pedal and accelerates up to a speed of 135 km/h. If it took 3.5 s to reach this speed, what is the average acceleration of the car?
Physics
1 answer:
mezya [45]3 years ago
8 0

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}

v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

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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