When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity
1 answer:
To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is
If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.
Therefore the correct answer is B.
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(a) Let's convert the final speed of the car in m/s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s: