Answer:
a. x= 83.03 m at t= 2 s
b. v=346.7 m/s at t= 4s
c. t=10.5s : maximum displacement occurs
d. t= 7s : maximum velocity
Explanation:
Definitions
acceleration :a(t) = dv/dt :Derived from velocity with respect to time
Velocity : V(t)=dx/dt : Derived from Displacement: with respect to time
Displacement: X(t)
Developing of problem
we have a(t) =10 (-t + 2 ) (t-5)+ 100
a(t) =(-10 t + 20) (t-5)+ 100 = -10 t²+50t+ 20t-100+ 100=--10 t²+70t
a(t)=--10 t²+70t Equation (1)
a(t) = dv/dt
-10 t²+70t=dv/dt
dv=(-10 t²+70t)dt
We apply integrals to both sides of the equation
∫dv=∫(-10 t²+70t)dt
[tex]v=\frac{-10t^{3} }{3} +\frac{70t^{2} }{2} +C_{1}[/tex]
[tex]v=\frac{-10t}{3} +35t^{2} +C_{1}[/tex]
at time t=0, v=0, then, C₁=0
[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] Equation (2)
[tex]v=\frac{dx}{dt}[/tex]
dx= vdt
We apply integrals to both sides of the equation and we replace v(t) of the equation (2)
[tex]\int\limits {\, dx =\int\limits{\frac{(-10t^{3} }{3}+35t^{2} ) } \, dt[/tex]
[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +C_{2}[/tex]
at time t=0, x=3, then,C₂=3
[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +3}[/tex] Equation (3)
a)Displacement at t=2s
We replace t=2s in the Equation (3)
[tex]x=\frac{-5(2)^{4} }{6} +\frac{35(2)^{3} }{3} +3}[/tex]
x= 83.03 m at t= 2 s
b) Velocity at t=4s
We replace t=4s in the Equation (2)
[tex]v=\frac{-10(4)^{3}}{3} +35(4)^{2}}[/tex]
v=346.7 m/s at t= 4s
c)Time at which maximum displacement occurs
at maximum displacement :v= [tex]\frac{dx}{dt} =0[/tex]
in the Equation (2) :
[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] =0
[tex]35t^{2} } =\frac{10t}{3}[/tex] ,we divide by t on both sides of the equation
[tex]t=\frac{35*3}{10}[/tex]
t=10.5s
d. maximum velocity over the interval
at maximum velocity :a= [tex]\frac{dv}{dt} =0[/tex]
in the Equation (1)
-10 t²+70t = 0 we multiply the equation by -1 and factor
10t ( t-7) =0
t= 7 s
