Suppose you see a duck flying north at 30 km/h. To another duck flying alongside, the first duck seems to be stationary. In other words, the velocity of a particle depends on the reference frame of whoever is observing or measuring the velocity. For our purposes, a reference frame is the physical object to which we attach our coordinate system. In everyday life, that object is the ground. For example, the speed listed on a speeding ticket is always measured relative to the ground. The speed relative to the police officer would be different if the officer were moving while making the speed measurement.
Suppose that Alex (at the origin of frame A in Fig. 4-20) is parked by the side of a highway, watching car P (the “particle”) speed past. Barbara (at the origin of frame B) is driving along the highway at constant speed and is also watching car P. Suppose that they both measure the position of the car at a given moment. From Fig. 4-20 we see that
The equation is read: “The coordinate xPA of P as measured by A is equal to the coordinate xPB of P as measured by B plus the coordinate xBA of B as measured by A.” Note how this reading is supported by the sequence of the subscripts.
Taking the time derivative of Eq. 4-40, we obtain
or (because v = dx/dt)
This equation is read: “The velocity vPA of P as measured by A is equal to the velocity vPB of P as measured by B plus the velocity vBA of B as measured by A.” The term vBA is the velocity of frame B relative to frame A. (Because the motions are along a single axis, we can use velocity components along that axis in Eq. 4-41 and omit overhead vector arrows.)
Here we consider only frames that move at constant velocity relative to each other. In our example, this means that Barbara (frame B) drives always at constant velocity vBA relative to Alex (frame A). Car P (the moving particle), however, can change speed and direction (that is, it can accelerate).
To relate an acceleration of P as measured by Barbara and by Alex, we take the time derivative of Eq. 4-41:
Because vBA is constant, the last term is zero and we have
In other words,
Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle.
CHECKPOINT 7 The table here gives velocities (km/h) for Barbara and car P of Fig. 4-20 for three situations. For each, what is the missing value and how is the distance between Barbara and car P changing?
In Fig. 4-20, suppose that Barbara’s velocity relative to Alex is a constant vBA = 52 km/h and car P is moving in the negative direction of the x axis.
(a) If Alex measures a constant velocity vPA = −78 km/h for car P, what velocity vPB will Barbara measure?
Solution: The Key Idea here is that we can attach a frame of reference A to Alex and a frame of reference B to Barbara. Further, because these two frames move at constant velocity relative to each other along one axis, we can use Eq. 4-41 to relate vPB to vPA and vBA:
If car P were connected to Barbara’s car by a cord wound on a spool, the cord would be unwinding at a speed of 130 km/h as the two cars separated.
(b) If car P brakes to a stop relative to Alex (and thus relative to the ground) in time t = 10 s at constant acceleration, what is its acceleration aPA relative to Alex?
Solution: The Key Idea here is that, to calculate the acceleration of car P relative to Alex, we must use the car’s velocities relative to Alex. Because the acceleration is constant, we can use Eq. 2-11 (v = v0 + at) to relate the acceleration to the initial and final velocities of P. The initial velocity of P relative to Alex is vPA = −78 km/h and the final velocity is 0. Thus, Eq. 2-11 gives us
(c) What is the acceleration aPB of car P relative to Barbara during the braking?
Solution: The Key Idea is that now, to calculate the acceleration of car P relative to Barbara, we must use the car’s velocities relative to Barbara. We know the initial velocity of P relative to Barbara from part (a) (vPB = −130 km/h). The final velocity of P relative to Barbara is −52 km/h (this is the velocity of the stopped car relative to the moving Barbara). Thus, again from Eq. 2-11,
We should have foreseen this result: Because Alex and Barbara have a constant relative velocity, they must measure the same acceleration for the car.
Leave a Reply