At time t = 0 s, an object is observed at x = 0 m; and its position along the x axis follows this expression: x = -4t + t², where the units for distance and time are meters and seconds, respectively. What is the object's displacement $\Delta$
x between t = 1.0 s and t = 3.0 s?

Question

At time t = 0 s, an object is observed at x = 0 m; and its position along the x axis follows this expression: x = -4t + t², where the units for distance and time are meters and seconds, respectively. What is the object's displacement

\Delta

x between t = 1.0 s and t = 3.0 s?

Quizplus · Accepted Answer

The displacement is the change in position, so we need to find the difference in position between t = 3.0 s and t = 1.0 s:$$\Delta x = x(3.0 	ext{ s}) - x(1.0 	ext{ s})$$$$= (-4(3.0 	ext{ s}) + (3.0 	ext{ s})^2) - (-4(1.0 	ext{ s}) + (1.0 	ext{ s})^2)$$$$= -12 	ext{ m} + 9 	ext{ m} + 4 	ext{ m} - 1 	ext{ m}$$$$= -16 	ext{ m}$$Therefore, the displacement is -16 m, which is choice A.

At Time T = 0 S, an Object Is Observed Δ\DeltaΔ

At Time T = 0 S, an Object Is Observed $\Delta$