US Halliday, Fundamentals of Physics, 10e Physics (1201/1202/2110/2112/2113) The

Question

US Halliday, Fundamentals of Physics, 10e Physics (1201/1202/2110/2112/2113) The position of a particle as it moves along a y axis is given by y = ( 3.0cm)sin(nt / 4), with t in seconds and y in centimeters. (a) What is the average velocity of the particle between t0 and t 2.0 s? (b) What is the 2.0 s? (c) What is the average acceleration of the particlie between t-0 and t 2.0s? (d) What is the instantaneous acceleration of the particle at t0, 1.0, and 2.0s? velocity of the particle at to 0, 1.0, and tuOs t 2.0s Units tmOs t-1.0s t 2.0s nits (d) Number Question Attempts: of 12 used savt ro. LATER

Explanation / Answer

Given

position of the particle is  

y(t) = 3.0 cm sin (pi*t/4)

a) we know that the average velocity is V avg = (y2-y1)/(t2-t1)

y1 is at time t=0 ==> y(1) = 3.0 sin (pi*0/4) = 0 cm

y2 is at time t=2 s ==> y(2) = 3.0 sin (pi*2/4) = 3 cm

now the average velocity is Vavg = (3-0)/(2-0) cm/s = 1.5 cm/s

b) instantaneous velocity of the particle at t= 0 s is  

v(t) = dy(t)/dt = 3*pi cos(pi*t/4)

now v(0) = 3*pi cos(pi*0/4) = 9.425 cm/s

v(1) = 3*pi cos(pi*1/4) = 6.664 cm/s

and v(2) = 3*pi cos(pi*2/4) = 0 cm/s

c) average acceleration is a_avg = (v2-v1) /(t2-t1)

a(1) = dv(1)/dt

from above instantaneous velocities at t=0 s , velocity is V(0) = 9.425 cm/s  

and the velocity at time t= 2s is v(2) = 0 cm/s

now the average acceleration is v_avg = (0-9.425)/(2-0) cm/s2 = -4.7125 cm/s2

d) the instantaneous acceleration is a(t) = dv(t)/dt = -3*pi^2 sin(pi*t/4)

at

t= 0 s is , a(0) = -3*pi^2 sin(pi*0/4) = 0 cm/s2

at t= 1 s is , a(0) = -3*pi^2 sin(pi*1/4) = -20.937 cm/s2

at t= 2 s is , a(0) = -3*pi^2 sin(pi*0/4) = -29.609 cm/s2

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