Question

the graph below shows ( v_x ) versus ( t ) for an object moving along a straight line. how would you find the acceleration ( a_x ) at ( t = 7.0 , \text{s} )?

multiple choice

• compute the area under the graph between ( t = 0.0 , \text{s} ) and ( t = 14 , \text{s} ).
• take the slope of the graph between ( t = 5.0 , \text{s} ) and ( t = 9.0 , \text{s} ).
• compute the area under the graph between ( t = 5.0 , \text{s} ) and ( t = 9.0 , \text{s} ).
• take the average of the slopes of the graph between ( t = 0.0 , \text{s} ) and ( t = 14 , \text{s} ).

Explanation:

Acceleration \( a_x \) in a velocity - time (\( v_x - t \)) graph is the slope of the \( v_x - t \) graph (\( a_x=\frac{\Delta v_x}{\Delta t} \)). At \( t = 7.0 \, \text{s} \), it lies between \( t = 5.0 \, \text{s} \) and \( t = 9.0 \, \text{s} \). So, to find \( a_x \) at \( t = 7.0 \, \text{s} \), we take the slope of the graph in this time interval as the graph is linear (or we can approximate the slope) in this region. The area under the \( v - t \) graph gives displacement, not acceleration, so options involving area calculation are incorrect. Taking the average of slopes over a large interval (0 - 14 s) or area - related calculations don't give acceleration at a specific time like \( t = 7.0 \, \text{s} \).

Answer:

Take the slope of the graph between \( t = 5.0 \, \text{s} \) and \( t = 9.0 \, \text{s} \).