The labeled graphs below represent the upward trajectories of 4 bodies, two slid

Question

The labeled graphs below represent the upward trajectories of 4 bodies, two sliding upwards on frictionless inclines and two in free flight. Note that all 4 bodies reach the same maximum height (9 meters) after traveling from the same initial elevation (0 m).

• The initial speed of P is (greater than less than equal to) that of Q.
• The initial speed of N is (greater than less than equal to) that of Q.
• The time of travel of R is (greater than less than equal to) that of Q.
• The time of travel of P is (greater than less than equal to) that of Q
• The initial speed of R is (greater than less than equal to) that of N.
• The time of travel of N is (greater than less than equal to) that of R

Explanation / Answer

1.equal to ,Because they have the same average velocity (as each experiences uniform acceleration) but Q has further to go

2. less than ,They have the same vertical velocity component but N has a smaller horizontal component as explained above. So the initial speed (magnitude of initial velocity) of N is smaller than R's.

3. equal to ,Explained above.

4.Will be same (due to same vertical velocity)

5.Speed R< Speed Q (final velocity will be zero for R so initial velocity will be less)

6. less than

The time of travel for N is the time to reach max height. Considering vertical motion only, since h = ½gt², t = [(2h/g].

The time of travel of Q can be calculated using height change and the vertical component of acceleration; this will be gsin() where is the angle of inclination to the horizontal. So t = [2h/(gsin())]

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