Question

Please really need help with question d really don't understand

Answer 1

Answer:

$\textsf{a)} \quad v=12t^2-6t-18$

$\textsf{b)} \quad a=24t-6$

c)   1.5 s

d)  -19.25 m

e)   24.5 m

Step-by-step explanation:

Displacement

$x=4t^3-3t^2-18t+1$

(where t ≥ 0 and x is in meters)

Part (a)

To find the equation for velocity, differentiate the equation for displacement:

$\implies v=\dfrac{\text{d}x}{\text{d}t}=12t^2-6t-18$

(where t ≥ 0 and v is in meters per second)

Part (b)

To find the equation for acceleration, differentiate the equation for velocity:

$\implies a=\dfrac{\text{d}v}{\text{d}t}=24t-6$

(where t ≥ 0 and a is in meters per second squared)

Part (c)

The particle comes to rest when its velocity is zero:

$\begin{aligned}v & = 0\\\implies 12t^2-6t-18 & = 0\\6(2t^2-t-3) & = 0\\2t^2-t-3 & = 0\\2t^2-3t+2t-3 & = 0\\t(2t-3)+1(2t-3) & = 0\\(t+1)(2t-3) & = 0\\\implies t & =-1, \dfrac{3}{2}\end{aligned}$

As t ≥ 0, the particle comes to rest at 1.5 s.

Part (d)

Substitute the found value of t from part (c) into the equation for displacement to find where the particle comes to rest:

$\implies 4(1.5)^3-3(1.5)^2-18(1.5)+1=-19.25\: \sf m$

Part (e)

We have determined that the particle is at rest at 1.5 s.

Therefore, to find how far the particle traveled in the first 2 seconds, we need to divide the journey into two parts: before and after it was at rest.  

The first leg of the journey is the first 1.5 s and the second leg of the journey is the next 0.5 s.

At the beginning of the journey, t = 0 s.

$\textsf{when }t=0: \quad x=4(0)^3-3(0)^2-18(0)+1=1$

Therefore, when t = 0, x = 1

When t = 1.5 s, x = -19.25 m (from part (d)).

⇒ Total distance traveled in first 1.5 s = 1 + 19.25 = 20.25 m

When t = 2, x = -15 m.

Therefore, the particle has traveled 19.25 - 15 = 4.25 m in the last 0.5 s of its journey.

Total distance traveled:

1 + 19.25 + 4.25 = 24.5 m

Refer to the attached diagram

When the particle is at rest, it changes direction.  If we model its journey using the x-axis, for the first leg of its journey (0 - 1.5 s) it travels in the negative direction (to the left). At 1.5 s it stops, then changes direction and travels in the positive direction (to the right), arriving at -15 at 2 seconds.