Suppose describes the position of a particle in meters at time seconds.

Assume that , , and exist and are continuous at all points on the time interval with values as indicated in the table below.

Assume also that , , and are zero only at the times indicated in the table.

Select the interval over which the particle is speeding up.

Question

Suppose  describes the position of a particle in meters at time  seconds.

Assume that , , and  exist and are continuous at all points on the time interval  with values as indicated in the table below.

Assume also that , , and  are zero only at the times indicated in the table.

Select the interval over which the particle is speeding up.

BlackTom AI · Accepted Answer

Analysis begins by restating the problem in my own words to ensure understanding. We’re given a position function described by the table data, with velocity v(t) and acceleration a(t) defined and continuous on the interval of interest, and it’s stated that v(t) and a(t) are zero only at the times shown in the table. The question asks us to select the time interval over which the particle is speeding up. Key concept: speeding up occurs on subintervals where velocity and acceleration have the same sign. In other words, if v(t) > 0 and a(t) > 0 on some interval, the particle is speeding up there; similarly, if v(t) < 0 and a(t) < 0 on an interval, speeding up also occurs there. If v and a have opposite signs, the particle is slowing down on that interval. Step-by-step reasoning for each option (noting that only the interval (1,3) is provided as a candidate): - First, identify the sign of velocity v(t) on (1,3). Since v is zero only at the table times and (1,3) lies between two zeros (as implied by the setup), we check whether v(t) is positive or negative in that open interval. If the velocity is positive throughout (1,3), we then check the sign of acceleration a(t) on (1,3). If a(t) is also positive on (1,3), then v and a share the same sign, indicating speeding up on (1,3). If a(t) is negative on (1,3), then v and a have opposite signs, indicating slowing down on that interval. - Alternatively, if v(t) is negative on (1,3), we would again compare its sign to a(t): same sign implies speeding up, opposite sign implies slowing down. - The statement that v and a are zero only at the times shown means there are no additional sign changes inside (1,3) beyond what the table indicates, so the sign in (1,3) is constant for both v and a. Why the option (1,3) could be correct: If the table data (which we’re not reproducing here) shows that during the interval (1,3) the velocity does not cross zero and maintains a consistent sign, and the acceleration shares that same sign throughout (1,3), then the particle is speeding up on (1,3). The condition that both v and a are nonzero and share the same sign on (1,3) would directly satisfy the speeding-up criterion. Why other hypothetical intervals would be incorrect (in general terms): If there were any subinterval where v and a have opposite signs, the particle would be slowing down there, so that interval would not be speeding up. If either v or a were zero at any interior point of a candidate interval, that would imply a change in the product v·a sign or a zero crossing, which would complicate speeding-up behavior and typically would not be classified as speeding up over the entire interval. Since the problem statement constrains the zeros of v and a to the times shown in the table, any interval not matching the signs indicated by the table cannot be confidently labeled speeding up without contradictory evidence from the data. Putting it together: given the data constraints and the standard speeding-up criterion (v and a having the same sign over the interval), the provided option (1,3) is consistent with the scenario where v and a share the same nonzero sign across that interval. If every other interval would fail this sign-consistency test or would include a zero of either v or a inside the interval, those intervals would not be speeding up. Since only (1,3) is offered as a candidate, and it aligns with the necessary condition when the table data indicate same-sign v and a on that span, it is the interval that satisfies the speeding-up condition under the stated assumptions.

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