Question

A particle moves along a straight line so that its velocity, v ms^-1, at time t seconds is given by v=240+20t-10t^2, for 0≤t≤6.

Answer 1

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.

Answer 2

The speed will be greatest on the Vextes of equation

So,

Xv = - b / 2a

As b = 20 and a = -10

Xv = - 20 / 2 . -10

Xv = 20 / 20

Xv = 1 s

Now replace Xv on the equation to find the value of speed

V(t) = 240 + 20t -10t^2

V(Xv) = Vmaximum

V(1) = 240 + 20.(1) - 10.(1)^2

V(1) = 240 + 20 -10

V(1) = 230m/s
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2:

If the speed is zero, then, there isn't acceleration.

Acceleration also is zero.