A cat walks in a straight line, which we shall call the x-axis, with the positiv

Question

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). Find the cat's velocity at t = 4.0 s and at t = 7.0 s What is the cat's acceleration at t = 3.0 s? At t = 6.0 s? At t = 7.0s? What distance does the cat move during the first 4.5 s? From t = 0 to t = 7.5 s? Assuming that the cat started at the origin, sketch clear graphs of the cat's acceleration and position as functions of time.

Explanation / Answer

The cat has starated with an initial velocoty (u) of 8 cm/s and has a decreasing velocity with time. That is it has negative acceleration (deceleration) of 8/6 = 4/3 cm /s2. so th answers respectively are (a) velocity a 4s is given by the formula v=u+at i.e. 8-(4/3)4=8/3 cm/s or 2.666 cm/s

cats velocity at 7s. the same formula when applied v=8-(4/3)7=-4/3 cm/s or -1.33cm/s i.e. it is now going back ward.

(b) Since the graph is a straight line one the acceleration is uniform and hence the acceleration is the same at all the three times i.e. is -4/3 or -1.333 cm/s

(c) the distance moved in the first 4.5 s is given by the formula S= ut+1/2at2 =(8x4.5) - 0.5x4/3x(4.5)2 =36-13.5 = 22.5 cm

from t=0 to t= 7.5 sec, the distance moved

at t=0 the cat is at s=vt =8x6=48 cm from the origin on the x axis.

at t=7.5 s the cat is (according to the formula S=ut+1/2at2 i.e. at 22.5 cm from the origin

so it had moved back by 25.5 cm (d) the graph of its acceleration with time is constant in the negative axis. 4th quadrant as (grapg though drawn could not insert it here)

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