Question

A particle's position on the x-axis is given by the function (3t-4t+1) m a) Make a position-versus time graph for the interval 0< t <5 (time is measured in seconds) b) Determine the particle's velocity at t = 2 s c) Are there any turning points in the particle's motion? If so, in what position or positions? d) Where is the particle when Vx=8 m/s? e) Draw the velocity-versus time graph for the interval 0< t <5 (time is measured in seconds)

Answer 1

9 criven , y=(3+²44 +1 )m Q) First kind the valves set where y=0 O, SO, 3 t²-4ttl=0 of a t = ut 16-12 yI2 6 2x3 SO, t = ģ where yo Now find the maxima or minima in the plot, dy - O dt 3x2 t -4 = 0 t = 4 -0-67 and y(0.67) = -0.33 6 which is greater than zero thus dt² t=0.67 is a minimum. As there ú minimum point so it ☺ a parabola - li 23 6 3 Now dy 10 only one Now at a=o, we get get ya 1 .

Taking au all the information, we can plot y vs t oss ym) +3 1 + + + + + 3 + y → کہ -Y 1 -3 s A t cs) -1 -2 2 (6) V = dy 64-4 at at t = 289 23, V = 6X2-4 12 -4 = 8 m/s Thus V = 8 m/s at t = 26. to dear Here polynomice has a 1 turning its minima (9) A turning point is a point at conien the function change from inero increasing creasing or vise-versa mial has degree Thus there is 1 point ie which we calculated in calculated in part 'a'. The turning point is at A ie (you toy) = (0.67, -0.33) the point is (0.67,-0.33) (d) V = 8 m/s , so, so, dy at 8 > 6xt-4=8 SO, t = 2 y (2) = 3x4 = 3x4 - 4 X 2+1 12-8+= 5 Thus the particle į at when the x is 8 m/s. 5 m

e) m/s iu a is a linear equanon. at t=0, V = V = 6t-4 m/s - 4 m/s at v=0, 4 6 alo 2 3 0.67 S 0.67 Thus the and x-intere cept is y-intercept is -4

V in ५ (/s) 3 1 । - -- + + 0 + 3 1 ५ 2- 5 -2 -3 -4 -5 (5)7 -1 -2 .५