Question

velocity ——time
which statement best describes the motions shown in this graph?
(1 point)
if c and d have the same force acting on them, then d must have more mass.
if c and d have the same acceleration, then d must have a greater starting velocity.
if c and d have the same mass, then d must have a greater force acting on it.
if c and d have the same force acting on them, then d must have a greater acceleration.

Explanation:

Step1: Recall Newton's second law \(F = ma\) (where \(F\) is force, \(m\) is mass, and \(a\) is acceleration) and the concept of acceleration from a velocity - time graph.

The slope of a velocity - time graph gives the acceleration. Let's assume two objects \(C\) and \(D\). If we consider the formula \(a=\frac{F}{m}\) (derived from \(F = ma\)).

Step2: Analyze each option

• Option 1: If \(F_C=F_D\) and \(a_C > a_D\) (from the steeper slope of \(C\) in the velocity - time graph), then from \(m=\frac{F}{a}\), \(m_D>m_C\). But we need to check other options.
• Option 2: At \(t = 0\), both \(C\) and \(D\) have the same initial velocity (they start from the same point on the velocity - axis). So this option is wrong.
• Option 3: If \(m_C=m_D\), and \(a_C > a_D\) (slope of \(C\) is steeper), from \(F=ma\), \(F_C>F_D\). This option is correct.
• Option 4: Since the slope of \(C\) (acceleration of \(C\)) is steeper than the slope of \(D\) (acceleration of \(D\)), if \(F_C = F_D\), then \(a_C=\frac{F_C}{m_C}\) and \(a_D=\frac{F_D}{m_D}\). Since \(a_C > a_D\), \(m_D>m_C\) (not \(a_D>a_C\)). So this option is wrong.

Answer:

If \(C\) and \(D\) have the same mass, then \(D\) must have a greater force acting on it.