The graph below shows how the force, 𝐹, acting on a four-kilogram box changes over time. The box is initially at rest, and the direction of the force is constant. What is the acceleration of the box at 𝑡 equals five seconds? (a) One meter per second squared. (b) One-half meter per second squared. (c) Four meters per second squared. (d) 16 meters per second squared. (e) Two meters per second squared.
The question asks us to find the acceleration of a four-kilogram box at time 𝑡 equals five seconds. To help us find the answer, we’re given this graph here on the right with time 𝑡 in units of seconds on the horizontal axis and force 𝐹 in units of newtons on the vertical axis. Using this graph, we can find the force 𝐹 on the box at time 𝑡 equals five seconds. First, locate five seconds on the horizontal axis. That’s this point here halfway between four and six. Now, draw a vertical line parallel to the vertical axis from the point 𝑡 equals five seconds. Since all of the points on this line have a horizontal coordinate of five seconds, the point where this line intersects the graph, right here, is the point on the graph that corresponds to 𝑡 equals five seconds.
To find the value of the force here, we need to know the value of the vertical coordinate of this point. To find the value, we first draw a straight horizontal line parallel to the horizontal axis all the way to the vertical axis. All of the points on this line have the same vertical-axis coordinate. So, let’s examine this point right here on the vertical axis itself. This point is on the vertical axis right at the place marked four. So, its vertical-axis coordinate is four. But that means that all of the points on the horizontal line that we drew also have a vertical-axis coordinate of four, including the point at the intersection of the line for 𝑡 equals five and the graph of 𝐹.
So, we’ve discovered that the point on the graph that corresponds to 𝑡 equals five seconds has coordinates five comma four. In other words, the force that the box feels at time 𝑡 equals five seconds is four newtons. Now, we have information about the force on the box at the time of interest. The other information we have about the box is that its mass is four kilograms. Now, we know the force on the box and its mass, and this is enough information to extract the acceleration. All we need is Newton’s second law; force is equal to mass times acceleration. We know the force and we know the mass. So, all we need to do is plug in to get the acceleration.
Plugging in our values of force and mass, we have four newtons, the force, is equal to four kilograms, the mass, times the unknown acceleration. Let’s divide both sides by four kilograms. On the right-hand side, four kilograms divided by four kilograms is just one. On the left-hand side, we have four newtons divided by four kilograms. To help us do this division, let’s rewrite newtons in terms of kilograms, meters, and seconds. One newton is one kilogram meter per second squared. So, we can rewrite the left-hand side as four kilogram meters per second squared divided by four kilograms.
Now, we can clearly see that four kilograms divided by four kilograms is one, and we’re left with one meter per second squared. And this was the left-hand side of an equation whose right-hand side was acceleration. So, one meter per second squared is equal to the acceleration. Since the force we used in our calculation was the force on the box at five seconds, this is the acceleration of the box at five seconds. The acceleration of the box at 𝑡 equals five seconds is exactly what we’re looking for. So, this is the answer.
So, the acceleration at 𝑡 equals five seconds is one meter per second, and this is choice (a).