(II) The force on a bullet along the barrel of a firearm is gi...

Question

(II) The force on a bullet along the barrel of a firearm is given by the formula F = [740 ― (2.3 x 10⁵ s⁻¹ ) t] N over the time interval t = 0 to t = 3.0 x 10⁻³ s .

(c) Determine the impulse by integration.

(b) Use the graph to estimate the impulse given the bullet.

Accepted Answer

Everyone in this problem. The thrust F of a rocket engine as it burns, fuel is critical for its initial ascent. The thrust can be described by the equation F is equal to 2050 minus 1.25 multiplied by 10 to the exponent four multiplied by T that's measured in newtons where T is a time in seconds from ignition. The equation is valid from time zero to 0.16 seconds covering the critical phase of engine burn. We're asked to plot a graph of the thrust F versus time tt equals 0 to 16 seconds. And then use the graph to estimate the total impulse delivered by the rocket engine to the rocket during this interval by calculating the area under the curve. So we're given this lo graph, we have t time in seconds on the X axis, the force in newtons on the Y axis. We're gonna come back to that in just a minute to get that graph done. And then we're given four answer choices for the impulse option. A 160 Newton seconds. Option B 170 Newton seconds. Option C 180 Newton seconds and option D 190 Newton seconds. OK. So let's go back to where we have this blank graph and we want to plot the force that we were given. Now, if we take a look at that force that we were given, OK. We have F is 2050 minus some number multiplied by T. Well, this is just a linear equation in T, OK. So we expect that this is going to be a straight line. Now, the Y intercept of this line is 2050 the slope is negative 1.25 multiplied by 10 to the exponent four. So that we expect a straight line with a negative slope. Then what we can do is use a table of values to help us graph this. So we're gonna look at certain time points and we're gonna look at the force at those time points and we're gonna start at times zero and we're just gonna use the marks along the X axis. So we're gonna go 0.04 0.08 0.12 and 0.16. For each of these, we're gonna look at the force. So when time is zero, we get the force is 2050 minus 1.25 multiplied by 10 to the exponent four multiplied by zero. This just gives us 2050 for each of the next times we're gonna do the exact same thing plugging in T in the equation, whatever value we're using, I need to save a little bit of space. We're not gonna rewrite that each time. But when we do it for 0.04 we end up with 2050 minus 500 which gives us a force of 1550 newtons. When T is 0.08 we get 2050 minus 1000, which gives us 1050. When T is 0.12 we get 2050 minus 1500 which gives us 550. And finally, when T is 0.16 we get 2050 minus 2000, which gives us a force of 50 new ones. OK? So we can go to our plot, we're gonna put, put these points on, connect them with the straight line. So the 1st 1, 2051 time is zero. It's gonna be somewhere around here. Then we have 1550 20 0.4 somewhere around here. 1050 when T is 0.8 550 when T is 0.12. And finally 50 when T is 0.6163. And again, we can connect these with a nice straight line and that is going to be our curve or force time. Now, the question is asking us to estimate the impulse using this graph and recall that the impulse is gonna be the area under our force curve. Now, we can think of this area in two kind of segments. OK. So we have this rectangular segment that spans across from the force of 50. And so it spans at the bottom, we're gonna color that in green. And then for the rest of this, we have this triangular segment that covers the rest of the area underneath the curve. OK. And so our impulse, what we can call Jay, it's gonna be the area under the curve which is just gonna be equal to the area of the rectangular region, who was the area of this triangular region. Now, the area of a rectangle is just the length times the width in the area of a triangle is one half multiplied by base multiplied by height. OK? So if we look at this green little rectangle, we think about the length of it and the width of it and we have a height of 50 newtons and it goes across from zero to 0.16 seconds. And so this width is 0.16 seconds. So what we get here is 50 noons multiplied by 0.16 seconds for that area, add the area of the triangle. Now, the base of the triangle is just like the width of that rectangle. It's going from zero to 0.16 seconds. So the base is gonna be 0.16 seconds. You think about the height of the triangle? It's going from 50 noons. Ok. Be careful. It's not starting at zero. It's starting at 50 noons and going up to 2050 noons. And so that height 2050 minus 50 is gonna be 2000. That distance between those two points is 2000. And so we have one half multiplied by 0.16 seconds, multiplied by 2000 noons. When we work this out, we get that, our impulse is gonna be eight Newton seconds. What? 160 noon seconds, which gives us an impulse of 168 noon seconds. Now, this question was asking us to approximate and to estimate the total impulse for estimating, we're gonna choose the closest value we're looking at kind of multiples of 10 here. And so the closest estimate for the impulse given the graph that we have is gonna be option B 170 Newton seconds. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

(II) The force on a bullet along the barrel of a firearm is given by the formula F = [740 ― (2.3 x 10⁵ s⁻¹ ) t] N over the time interval t = 0 to t = 3.0 x 10⁻³ s . Plot a graph of F versus t for t = 0 to t = 3.0 ms .
(b) Use the graph to estimate the impulse given the bullet.

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Struggling with Physics?

(II) The force on a bullet along the barrel of a firearm is given by the formula F = [740 ― (2.3 x 10⁵ s⁻¹ ) t] N over the time interval t = 0 to t = 3.0 x 10⁻³ s . Plot a graph of F versus t for t = 0 to t = 3.0 ms .(b) Use the graph to estimate the impulse given the bullet.

Watch next

Impulse with Variable Forces practice set

(II) The force on a bullet along the barrel of a firearm is given by the formula F = [740 ― (2.3 x 10⁵ s⁻¹ ) t] N over the time interval t = 0 to t = 3.0 x 10⁻³ s . Plot a graph of F versus t for t = 0 to t = 3.0 ms .
(b) Use the graph to estimate the impulse given the bullet.