Answer:
(i) The speed of the particle is greatest at t=1 and maximum speed is 500 m/s.
(ii) The acceleration of the particle is -100 when its speed is zero.
Step-by-step explanation:
(i)
We need to find the the time at which time at which the speed of the particle is greatest. It means we need to find the value of t at which velocity function is maximum.
If a function is \(f(x)=ax^2+bx+c\), then the vertex of the function is
\((-\frac{b}{2a},f(-\frac{b}{2a}))\)
The given velocity function is
\(v=240+20t-10t^2\)
For 0≤t≤6.
Here, a=-10, b=20 and c=240.
The leading coefficient is negative it means the vertex of the velocity function is the point of maxima.
\(-\frac{b}{2a}=-\frac{20}{2(-10)}\Rightarrow -\frac{20}{-20}=1\)
At t=1,
\(v=240+20(1)-10(1)^2=250\)
Therefore, the speed of the particle is greatest at t=1 and maximum speed is 500 m/s.
(ii)
Equate the velocity function equal to zero, to find the time at which speed is zero.
\(v=0\)
\(240+20t-10t^2=0\)
\(10(24+2t-t^2)=0\)
Splitting the middle term we get
\(10(24+6t-4t-t^2)=0\)
\(10(6(4+t)-t(4+t)=0\)
\(10(6-t)(4+t)=0\)
Using zero product property,
\((6-t)=0\Rightarrow t=6\)
\(4+t=0\Rightarrow t=-4\)
Time can not be negative. So, at t=6 the speed is zero.
Differentiate the velocity function with respect to t.
\(a(t)=v'=20(1)-10(2t)\)
\(a(t)=20-20t\)
Substitute t=6 in the above function.
\(a(t)=20-20(6)\)
\(a(t)=-100\)
Therefore the acceleration of the particle is -100 when its speed is zero.
