Answer:
c) 1.5 s
d) -19.25 m
e) 24.5 m
Step-by-step explanation:
<u>Displacement</u>
(where t ≥ 0 and x is in meters)
<h3><u>Part (a)</u></h3>
To find the equation for <u>velocity</u>, differentiate the equation for displacement:
(where t ≥ 0 and v is in meters per second)
<h3><u>Part (b)</u></h3>
To find the equation for <u>acceleration</u>, differentiate the equation for velocity:
(where t ≥ 0 and a is in meters per second squared)
<h3><u>Part (c)</u></h3>
The particle comes to <u>rest</u> when its velocity is zero:
As t ≥ 0, the particle comes to rest at 1.5 s.
<h3><u>Part (d)</u></h3>
<u>Substitute</u> the found value of t from part (c) into the equation for displacement to find where the particle comes to rest:
<h3><u>Part (e)</u></h3>
We have determined that the <u>particle is at rest at 1.5 s</u>.
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.
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.
<u>Total distance traveled</u>:
1 + 19.25 + 4.25 = 24.5 m
<u>Refer to the attached diagram</u>
When the particle is at rest, it <u>changes direction</u>. 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.