At #t=0#, a particle is located at #x=25m# and has a velocity of #15m/s# in the positive #x# direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at #t=5.0s#?

http://www.laphysics.com/physicsap/tests/kinematics/kinematicspractice.pdf

1 Answer
May 27, 2016

Answer:

#"the position of the particle at t=5 is "25/6" m"#

Explanation:

#"using diagram"#
#a(t)=a_i-k*t#

#a_i=6#
#k=tan alpha=6/6=1#

#a(t)=6-t#

#v(t)=int (6-t)d t#

#v(t)=6t-1/2t^2+C#

#"for t=0 v=15 m/s C=15"#

#v(t)=6t-1/2t^2+15#

#x(t)=int v(t) d t=int (6t-1/2 t^2+15) d t#

#x(t)=1/2*6* t^2-1/2*1/3*t^3+15t+C#

#x(t)=3t^2-1/6t^3+15t+C#

#"for t=0 x=25 m, C=25#

#x(t)=3t^2-1/6x^3+15t+25#

#x(5)=3*5^2-1/6* 5^3+15*5+25#

#x(5)=75-125/6+75+25#

#x(5)=25-125/6#

#x(5)=(150-125)/6#

#x(5)=25/6#