Question

2. A mass moves along a straight line, which we will call the x-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t) = 40.00cm + 16.00 en/s t-10.00cm/s2 t2 A. Find the object's initial velocity, initial position, and initial acceleration. B. At what time t is the velocity of the object zero? C. How long after starting does it take the object to return to its starting point? D. At what times t is the object a distance of 8.0 cm from its starting point? What is the velocity (magnitude and direction) of the mass at each of those times?

Answer 1

A) intial velocity = d/dt(x(t))= 16- 2*10t

put t=0, initial velocity= 16cm/s

intial distance (put t=0 in given equation) =40 + 16*0 - 10*0² = 40 cm

initial accelaration = d/dt(initial velocity) = d/dt(16- 2*10t) = -20 m/s²

B) differentiate given equation to find velocity at a certain time

d/dt(x(t))= 16 - 2*10*t

put d/dt(x(t))= 0 that is

16- 2*10*t =0

t = 16/20 = 0.8 sec

C) put x(t)=40 in above equation,

40 = 40 +16*t -10*t²

solving for t we get

16*t = 10*t²

hence t= 0sec and t= 1.6 sec

so it will return at its original place at t= 1.6sec

D) put x(t) =8 in above equation and solve for t that is

8 = 40 +16*t -10*t²

t = 2.75 sec

for velocity differentiate x(t) that is we get

velocity = 16 - 2*10t

put t= 2.75 sec

velocity = 16- 2*10*2.75 = -39 cm/s