My Math Forum Kinematics Question

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 October 13th, 2016, 06:29 AM #1 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 Kinematics Question Link for graph: https://postimg.org/image/9yefgd69p/ The above graph represents the motion of two vehicles: A and B a. What is the acceleration of vehicle A during the first 5.0 s? b. Calculate the distance between vehicle A and B after 6.0 s? c. What is the instantaneous acceleration of vehicle A at the 10.0 s mark? d. What is the average velocity of vehicle A from start to 11.0 s? e. Sketch an acceleration vs time graph for vehicle A. (rough sketch only showing the general shape) Last edited by skipjack; October 13th, 2016 at 06:44 AM.
 October 13th, 2016, 06:36 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions Have you tried the question yet? What have you got? Thanks from skeeter and topsquark
 October 13th, 2016, 07:49 AM #3 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 @Benit13 I solved part a) and part b) For part a), I got 5 m/s^2 using the formula a=(v-u)/t and for part b) I know that the distance by vehicle A is 0.5x5x25+(6-5)(25)=87.5m and the distance by vehicle B is 0.5x(15+20)x6=105m so, I calculated the distance between the two vehicles = 105m-87.5m=17.5m , for parts c) and d) I didnt know how to find the instantaneous acceleration of vehicle A at the 10.0 s mark and find the average velocity of vehicle A from start to 11.0 s. for part e) I sketched this graph but I'm not sure about it here is the link for the sketch I made: https://postimg.org/image/55zj9cknd/
 October 13th, 2016, 08:42 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions a) and b) are correct. For part c) the acceleration is calculated using the same formula as part a) a = (v-u)/t except that you need to draw your own triangle. To do that, draw a tangent to the velocity curve at the point t=10 seconds and then use that tangent line as the hypotenuse of a small triangle. You can choose how big or small you want the triangle, but don't make it too small or the next step is harder! The next step is to divide the vertical length of the triangle (in m/s) by the horizontal length of the triangle (in seconds) to find out the acceleration. It's exactly the same thing you did in part a), except you're making the triangle yourself from a tangent rather than using the straight line that is already there. For part d) draw a big line between the start point (t=0 s) and the final point (t=11 s) and then do the same thing as part c) above, but for the big triangle. For part e)... yeah, your diagram is incorrect. The thing to remember is that if the velocity is changing at a constant rate the acceleration is constant (doesn't change). If the velocity doesn't change, the acceleration is zero. In other words, if there is a straight line on the velocity-time graph, the acceleration must be a straight horizontal line on an acceleration-time plot. Thanks from Vaptor Last edited by skipjack; October 13th, 2016 at 11:54 AM.
 October 14th, 2016, 03:13 AM #5 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 @Benit 13 for part c) i drew the tangent at t=10s and drew the triangle, I got (35m/s-25m/s)/2s=5m/s^2, is that the correct answer or have i done some mistake, for part d) i got (50m/s-0m/s)/11s=4.5m/s
 October 14th, 2016, 03:37 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions Your part d) is now correct. For part c), I would make sure you draw the tangent to the line at the point where t=10... If I draw the tangent, I get a point at t=11 seconds, 45 m/s and another point at t=9 seconds, 25 m/s. Therefore, when you draw the triangle with those two points, the hypotenuse touches the curve at t=10 seconds, 35 m/s. Therefore, the answer should be more like $\displaystyle a = \frac{45 - 25}{11 - 9} =\frac{20}{2} = 10 m/s^2$ Thanks from Vaptor
 October 14th, 2016, 07:35 AM #7 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 I need you to check one last thing for me please, for question e), i got this diagram after I followed the rules you gave me, I'm still not sure if its correct https://postimg.org/image/4sl2j9w3x/
 October 14th, 2016, 07:49 AM #8 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions That's wrong as well. Try answering these questions: i) What is the instantaneous acceleration of vehicle A at t = 1 s? ii) What is the instantaneous acceleration of vehicle A at t = 2 s? iii) What is the instantaneous acceleration of vehicle A at t = 3 s? iv) What is the instantaneous acceleration of vehicle A at t = 6 s? v) What is the instantaneous acceleration of vehicle A at t = 7 s? vi) What is the instantaneous acceleration of vehicle A at t = 8 s? Then put these points on the graph and draw it Thanks from Vaptor
 October 14th, 2016, 08:35 AM #9 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 What about now... https://postimg.org/image/dhgiw8au7/
 October 14th, 2016, 09:34 AM #10 Newbie   Joined: Aug 2016 From: Germany Posts: 7 Thanks: 0 What about now... https://postimg.org/image/dhgiw8au7/

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