An object is at x = 0 at t = 0 and moves along the x axis according to the veloc

Question

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph in Figure P2.62. What is the object's acceleration between 0 and 4.0 s? What is the object's acceleration between 4.0 s and 9.0 s? What is the object's acceleration between 13.0 s and 18.0 s? At what time(s) is the object moving with the lowest speed? At what time is the object farthest from x = 0? What is the final position x of the object at t = 18.0 s? Through what total distance has the object moved between t = 0 and t = 18.0 s?

Explanation / Answer

between 0 to 4 seconds, the veloicty is constant

hence acceleration=rate of change of velocity with repsect to time=0

part b:
between 4 seconds to 9 seconds, velocity increases linearly

hence acceleration=change in velocity/change in time

=(20-(-10))/(9-4)=6 m/s^2

part c:

between 13 seconds to 18 seconds, the velocity is decreasing linearly.

then acceleration=change in veloicty/change in time

=(0-20)/(18-13)=-4 m/s^2

part d:

lowest speed is -10 m/s^2 and it is between t=0 to t=4 seconds

part e:

equation for veloicty:

v=-10 m/s , 0<t<4

=>dx/dt=-10

==>x=-10*t+c

where c is a constant

at t=0, x=0==>c=0

at the end of t=4 seconds, x=-10*4=-40 m

v=6*t-34, for 4<t<9

==>dx/dt=6*t-34

==>x=3*t^2-34*t+c

at t=4 seconds, x=-40 m

==>-40=48-34*4+c

==>c=48

then x(t)=3*t^2-34*t+48

at t=9 seconds, x=-15 m

v=20 , 9<t<13 seconds

dx/dt=20

==>x=20*t+c

at t=9 seconds, x=-15 m

==>c=-195

hence x(t)=20*t-195 , 9<t<13

at x=13 seconds, x=65 m

for 13<t<18 , v=-4*t+72

==>dx/dt=-4t+72

==>x=-2*t^2+72*t+c

at t=13 seconds, x=65 m

==>c=-533

==>x(t)=-2*t^2+72*t-533

at t=18 seconds, x=115 m

hence maximum distance the object is from the origin is 115 m and at t=18 seconds

f)final position is 115 m

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