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.

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.

What distance does the cat move from t=0 to t=7.5s?

Answer: 26cm.

I have already exhausted the 9 tries allowed in this problem. I have no clue how to arrive at the correct answer displayed. The following is my attempt:

x = 0+8(7.5)+[(-4/3)(7.5^2)]/2 = 22.5cm

I don't understand. Do I need to split the equations from t: [0,6] and then t: [6,7.5] somehow? Please explain.

Answer 1

Eqyation of velocity of the line = v = -4/3t +8

put t= 7.5 we get v=-2 ;

total distance travelled = total area under the grah of v vs t :

= 0.5 * 8*6 + 0.5*2 *1.5 =25.5 cm is the answer !

Answer 2

here a ccelaration is constant a = V-u/t = (0-8)/6 = -1.33 cm/s^2

from V^2-U^2 = 2*a*s1

s1 = u*t+0.5*a*t^2 = 8*6-0.5*1.33*6^2= 24.06cm

from t = 6 to 7.5 s   the velocity is increasing so a =+ve

s2 = u*t+0.5*a*t^2 = 0+0.5*1.33*1.5^2 = 1.499 cm

total S = s1+s2 = 25.55 cm = 26 cm

Answer 3

a = dv/dt = -8/6 = -1.333m/s^2



displacement of the body from t = 0 to t = 6

s1 = u*t + 0.5*a*t^2

s1 = 24.004 cm m

displacement of the body from t = 6 to t = 7.5


s2 = 0.5*a*t^2 = -0.5*1.333*1.5^2 = -1.5 m

distance s2 = 1.5 m


total distance travelled = 24.004+1.5 = 25.5 m