Questions: Two objects A and B accelerate from rest with the same constant acceleration. Object A accelerates for twice as much time as object B, however. Which one of the following statements is true concerning these objects at the end of their respective periods of acceleration? A) Object A will travel twice as far as object B. B) Object A will travel four times as far as object B. C) Object A will travel eight times further than object B. D) Object A will be moving four times faster than object B. B) Object A will be moving eight times faster than object B.

Two objects A and B accelerate from rest with the same constant acceleration. Object A accelerates for twice as much time as object B, however. Which one of the following statements is true concerning these objects at the end of their respective periods of acceleration?
A) Object A will travel twice as far as object B.
B) Object A will travel four times as far as object B.
C) Object A will travel eight times further than object B.
D) Object A will be moving four times faster than object B.
B) Object A will be moving eight times faster than object B.

Transcript text: Two objects A and B accelerate from rest with the same constant acceleration. Object A accelerates for twice as much time as object B, however. Which one of the following statements is true concerning these objects at the end of their respective periods of acceleration? A) Object $A$ will travel twice as far as object $B$. B) Object A will travel four times as far as object B. C) Object A will travel eight times further than object B. D) Object $A$ will be moving four times faster than object $B$. B) Object A will be moving eight times faster than object B.

Solution

Solution Steps

Step 1: Understanding the Problem

We need to determine the relationship between the distances traveled and the final velocities of two objects, A and B, which start from rest and accelerate with the same constant acceleration. Object A accelerates for twice as much time as object B.

Step 2: Using Kinematic Equations

The kinematic equation for distance traveled under constant acceleration is: \[ s = ut + \frac{1}{2}at^2 \] Since both objects start from rest, $ u = 0 $, so the equation simplifies to: \[ s = \frac{1}{2}at^2 \]

Step 3: Calculating Distance Traveled

Let $ t_B $ be the time for which object B accelerates. Then, object A accelerates for $ 2t_B $.

For object B: \[ s_B = \frac{1}{2}a(t_B)^2 \]

For object A: \[ s_A = \frac{1}{2}a(2t_B)^2 = \frac{1}{2}a(4t_B^2) = 2a(t_B)^2 \]

Step 4: Comparing Distances

Comparing the distances: \[ \frac{s_A}{s_B} = \frac{2a(t_B)^2}{\frac{1}{2}a(t_B)^2} = 4 \] Thus, object A travels four times as far as object B.

Step 5: Calculating Final Velocities

The kinematic equation for final velocity under constant acceleration is: \[ v = u + at \] Since both objects start from rest, $ u = 0 $, so the equation simplifies to: \[ v = at \]

For object B: \[ v_B = a t_B \]

For object A: \[ v_A = a (2t_B) = 2a t_B \]

Step 6: Comparing Final Velocities

Comparing the final velocities: \[ \frac{v_A}{v_B} = \frac{2a t_B}{a t_B} = 2 \] Thus, object A will be moving twice as fast as object B.