Question

The figure below shows a graph of Vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. vy(m/s) 10 8 6 4 2 (s) 4 6 8 10 12 (a) Find the average acceleration for the time interval t = 0 to t = 6.0 s. m/s2 (b) Estimate the time at which the acceleration has its greatest positive value. s What is the value of the acceleration at that instant? m/s2 (c) When is the acceleration zero? Acceleration is zero when t = s and when t> s. (d) Estimate the maximum negative value of the acceleration. m/s2 At what time does it occur?

Answer 1

Average acceleration. (from t-o to t=6 AV a av At (8-0) m/s (6-0) s Sav = 1:33 m/s2 Ang

greatest Ang. 6 m/s 11.5 5 m/s2 4 s An e maximum b) upto t=45 the slope of Vx-t graph is creasin acceleration bas t=4s. positive value cat acceleration is acceleration will be When slope of Zero Vx-t graph is horizontal => t=65 A t) 10s And. d), At t=85 there negative acceleration because after This time blope mag opto de decreases. Zero Vx-t graph is a max 6 m/s -0.75 m/s2 8s And