Solution: y = a + bt + ct² − dt⁴ Velocity v = dy/dt = b + 2ct − 4dt³ Acceleration a = dv/dt = 2c − 12dt² At t = 0: v = b, acceleration = 2c. So the initial velocity and acceleration are b and 2c respectively.

Solution: For a semicircular path of radius 1.0 m, the actual DISPLACEMENT (straight-line distance from A to B) is the diameter, not the arc length: displacement = 2r = 2 × 1.0 = 2.0 m Average velocity = displacement / time = 2.0 m / 1.0 s = 2 m/s. (Average SPEED would use the arc length, πr ≈ 3.14 m, giving 3.14 m/s — that's a different quantity and matches a different option, but the question asks for average velocity.)

Solution: x = t³ − 6t² + 3t + 4 v = dx/dt = 3t² − 12t + 3 a = dv/dt = 6t − 12 Set a = 0: 6t − 12 = 0 → t = 2 s v at t = 2: 3(2)² − 12(2) + 3 = 12 − 24 + 3 = −9 m/s.

Solution: Average velocity magnitude = |displacement| / time, while average speed = total path length / time. Since the straight-line displacement between two points can never exceed the actual path length travelled, |average velocity| ≤ average speed always — the ratio is at most 1. The ratio equals 1 only for motion in a fixed direction with no reversal (straight-line, one-way motion); for any path with curvature or direction reversal it is strictly less than 1. Among the given choices, "always less than one" is intended as the general case (motion that isn't perfectly one-directional), though strictly the ratio can also equal one — the more complete statement would be "less than or equal to one."

Solution: x = a·e^(−αt) + b·e^(βt), with a, b, α, β all positive. v = dx/dt = −aα·e^(−αt) + bβ·e^(βt) dv/dt = aα²·e^(−αt) + bβ²·e^(βt) Both terms in dv/dt are always positive (since a, b, α, β, and the exponentials are all positive), so dv/dt > 0 for all t — velocity is always increasing. As t grows large, the −aα·e^(−αt) term decays to 0 while the bβ·e^(βt) term grows without bound, so v keeps increasing indefinitely — it never drops to zero or levels off.

Note: the question text stored for this item is incomplete — it cuts off right after "...is shown in [figure]" without stating what property is being asked about (velocity, acceleration, etc.). Based on the marked answer (E) and the shape of the curve: The curve rises to a peak at D (where the slope — velocity — is momentarily zero, since D is a local maximum of displacement), then falls before rising again toward F. Between D and F the curve changes concavity: concave-down near D, concave-up as it approaches the next rise. E sits at that inflection point, where the curvature (and therefore the second derivative — acceleration) crosses zero. So E is the point of zero acceleration, distinguishing it from D, where velocity (not acceleration) is momentarily zero.

Solution: The shape of a particle's trajectory is fixed by its initial velocity direction together with how the acceleration acts on it from then on. Speed alone (a scalar) carries no directional information and can't determine the path's shape. But velocity (giving the initial direction of motion) combined with acceleration (governing how that direction changes) together fully determine the path — e.g. constant acceleration plus a velocity not parallel to it produces a parabola (projectile motion); acceleration always perpendicular to velocity with constant speed produces a circle. So both velocity and acceleration are needed together.

Solution: A body at the top of a vertical throw, or at the extreme point of oscillation, has zero velocity at that instant but is still accelerating (e.g. under gravity, g downward) — so zero velocity does NOT force zero acceleration. That rules out "zero velocity necessarily zero acceleration." Uniform SPEED doesn't require uniform (constant-direction) acceleration — uniform circular motion has constant speed but continuously changing (centripetal) acceleration direction, so that option is false too. Non-uniform velocity generally means the velocity is changing, i.e. there IS acceleration — so "non-uniform velocity has zero acceleration" is false. That leaves: "a body having zero velocity will not necessarily have zero acceleration" — true, and the correct statement.

Solution: During the upward trip, air resistance (drag) opposes the (upward) motion, so it acts downward — same direction as gravity — giving a LARGER net deceleration than g alone. That brings the body to rest (v = 0) more quickly, so the time of ASCENT is shorter than it would be without air resistance. During the downward trip, drag opposes the (downward) motion, so it acts upward — opposing gravity — giving a SMALLER net acceleration than g. That makes the body fall more slowly, so the time of DESCENT is longer than it would be without air resistance. Comparing the two: the rise time (shortened by drag) is less than the fall time (lengthened by drag).

Solution: Let displacement x = k·t³ for some constant k. Velocity v = dx/dt = 3k·t² Acceleration a = dv/dt = 6k·t Since a is directly proportional to t, its magnitude grows steadily as time passes — it is increasing with time (not constant, and not zero for t > 0).

Solution: Let u be the launch speed and H the maximum height. Using v² = u² − 2gh: At the top (h = H), v = 0: 0 = u² − 2gH → u² = 2gH At half the height (h = H/2): v² = u² − 2g(H/2) = u² − gH = 2gH − gH = gH So v² = gH → H = v²/g = (10)² / 10 = 100/10 = 10 m. (Note: the marked option "10cm" appears to be a scraping/typo error for "10 m" — the computed value is 10 metres, off by a factor of 100 from what's printed. The physics and the numeric value 10 are correct; only the unit shown in the option looks wrong.)

Solution: Let the one-way distance be d. Time for X→Y at speed v_u is d/v_u; time for Y→X at speed v_d is d/v_d. Average speed = total distance / total time = 2d / (d/v_u + d/v_d) = 2 / (1/v_u + 1/v_d) = 2v_u v_d / (v_u + v_d). This is the harmonic mean of the two speeds — always the correct formula for a round trip covering equal distances at different constant speeds.

Solution: By symmetry of projectile motion, the last t seconds of the ASCENT (just before reaching the top) mirror the FIRST t seconds of a body released from rest and falling freely — because at the very top the velocity is momentarily zero, and running time backward from there looks exactly like free fall from rest. Distance fallen from rest in time t is (1/2)g t² — so the distance covered in the last t seconds of the ascent is also (1/2)g t², independent of the launch speed u.

Solution: Assertion: true — e.g. a ball at the highest point of its vertical throw has velocity = 0 but acceleration = g (still acting). Reason: also true as a general statement — a body is indeed momentarily at rest at the instant it reverses direction. But the Reason does not fully EXPLAIN the Assertion: there are cases with zero velocity and nonzero acceleration that have nothing to do with reversing direction — e.g. a body released from rest under gravity has v = 0 at t = 0 with acceleration g acting, and it never reverses direction at all. So the Reason describes only one specific scenario, not the complete/general explanation of the Assertion. Hence both statements are true, but the Reason is not the correct explanation of the Assertion.