A cat walks in a straight line, which we shall call the x-axis, with the positive direction to the right. As an observant physicist, you make measurements of this cat’s motion and construct a graph of the feline’s velocity as a function of time (?Fig. E2.30?). (a) Find the cat’s velocity at t = 4.0 s and at t = 7.0 s. (b) What is the cat’s acceleration at t = 3.0 s? At t = 6.0 s? At t = 7.0 s? (c) What distance does the cat move during the first 4.5 s? From t = 0 to t = 7.5 s? (d) Assuming that the cat started at the origin, sketch clear graphs of the cat’s acceleration and position as functions of time.

Solution 30E Step 1 of 6: (a) Find the cat’s velocity at t = 4.0 s and at t = 7.0 s. From the given v-t graph, Cat’s velocity at t= 4s is v= 2 cm/s Similarly at t= 7s, using straight line equation(y=mx+c), here v=mt-c Using slope,m= 1.33, c= 8 and t= 7s is v= -1.31 cm/s Step 2 of 6: (b) What is the cat’s acceleration at t = 3.0 s At t = 6.0 s At t = 7.0 s Cat’s acceleration can be calculated by the ratio of velocity by time, v a= t At t= 3 s, from graph v= 4 cm/s 2 2 Using 1 cm =10 m v= 4 × 10 m /s Using above equation, 4 ×10 m/s a= 3 s 2 2 a= 1.33 × 10 m/s At t= 6 s, from graph v= 0 cm/s Using above equation, a= 0 6 s a= 0 At t= 7 s, from graph v= -1.31 cm/s 2 2 Using 1 cm =10 m v= -1.31 × 10 m/s Using above equation, 1.31 ×10 m/s a= 3 s 2 2 a= - 0.43 × 10 m/s Step 3 of 6: (c) What distance does the cat move during the first 4.5 s From t = 0 to t = 7.5 s We know that, the area under the curve in the velocity -time graph will given the displacement. To calculate distance travelled for first 4.5 s Area= area of the trapezoid = a+bh 2 Using a=2, b=8 and h = 2.5 Displacement= area= 2+8 (2.5) 2 Displacement = 12.5 cm Therefore, the cat travels a distance of 12.5 cm in first 4.5 seconds.