WEBVTT
00:00:02.760 --> 00:00:07.980
The graph below shows how the force, 𝐹, acting on a four-kilogram box changes over time.
00:00:08.320 --> 00:00:11.950
The box is initially at rest, and the direction of the force is constant.
00:00:12.090 --> 00:00:15.000
What is the acceleration of the box at 𝑡 equals five seconds?
00:00:15.420 --> 00:00:17.450
(a) One meter per second squared.
00:00:17.870 --> 00:00:20.240
(b) One-half meter per second squared.
00:00:20.590 --> 00:00:22.650
(c) Four meters per second squared.
00:00:23.060 --> 00:00:25.240
(d) 16 meters per second squared.
00:00:25.710 --> 00:00:27.720
(e) Two meters per second squared.
00:00:28.370 --> 00:00:34.340
The question asks us to find the acceleration of a four-kilogram box at time 𝑡 equals five seconds.
00:00:34.600 --> 00:00:44.470
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.
00:00:44.770 --> 00:00:49.510
Using this graph, we can find the force 𝐹 on the box at time 𝑡 equals five seconds.
00:00:49.840 --> 00:00:52.680
First, locate five seconds on the horizontal axis.
00:00:52.990 --> 00:00:55.830
That’s this point here halfway between four and six.
00:00:56.370 --> 00:01:01.190
Now, draw a vertical line parallel to the vertical axis from the point 𝑡 equals five seconds.
00:01:01.480 --> 00:01:12.780
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.
00:01:13.000 --> 00:01:17.360
To find the value of the force here, we need to know the value of the vertical coordinate of this point.
00:01:17.790 --> 00:01:24.430
To find the value, we first draw a straight horizontal line parallel to the horizontal axis all the way to the vertical axis.
00:01:24.900 --> 00:01:27.870
All of the points on this line have the same vertical-axis coordinate.
00:01:28.300 --> 00:01:31.860
So, let’s examine this point right here on the vertical axis itself.
00:01:32.380 --> 00:01:35.410
This point is on the vertical axis right at the place marked four.
00:01:35.660 --> 00:01:37.710
So, its vertical-axis coordinate is four.
00:01:38.210 --> 00:01:49.100
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.
00:01:49.420 --> 00:01:55.470
So, we’ve discovered that the point on the graph that corresponds to 𝑡 equals five seconds has coordinates five comma four.
00:01:56.180 --> 00:02:01.190
In other words, the force that the box feels at time 𝑡 equals five seconds is four newtons.
00:02:01.790 --> 00:02:05.480
Now, we have information about the force on the box at the time of interest.
00:02:05.900 --> 00:02:09.560
The other information we have about the box is that its mass is four kilograms.
00:02:10.080 --> 00:02:15.120
Now, we know the force on the box and its mass, and this is enough information to extract the acceleration.
00:02:15.580 --> 00:02:20.700
All we need is Newton’s second law; force is equal to mass times acceleration.
00:02:21.190 --> 00:02:23.480
We know the force and we know the mass.
00:02:23.620 --> 00:02:26.380
So, all we need to do is plug in to get the acceleration.
00:02:26.850 --> 00:02:34.570
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.
00:02:35.280 --> 00:02:37.330
Let’s divide both sides by four kilograms.
00:02:37.520 --> 00:02:40.870
On the right-hand side, four kilograms divided by four kilograms is just one.
00:02:42.550 --> 00:02:46.060
On the left-hand side, we have four newtons divided by four kilograms.
00:02:46.280 --> 00:02:51.050
To help us do this division, let’s rewrite newtons in terms of kilograms, meters, and seconds.
00:02:51.410 --> 00:02:54.350
One newton is one kilogram meter per second squared.
00:02:54.770 --> 00:03:00.360
So, we can rewrite the left-hand side as four kilogram meters per second squared divided by four kilograms.
00:03:00.690 --> 00:03:06.630
Now, we can clearly see that four kilograms divided by four kilograms is one, and we’re left with one meter per second squared.
00:03:07.050 --> 00:03:11.150
And this was the left-hand side of an equation whose right-hand side was acceleration.
00:03:11.390 --> 00:03:14.230
So, one meter per second squared is equal to the acceleration.
00:03:14.820 --> 00:03:21.370
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.
00:03:21.640 --> 00:03:25.310
The acceleration of the box at 𝑡 equals five seconds is exactly what we’re looking for.
00:03:25.610 --> 00:03:26.560
So, this is the answer.
00:03:27.030 --> 00:03:31.420
So, the acceleration at 𝑡 equals five seconds is one meter per second, and this is choice (a).