<h2>
Answer:</h2>
velocity = 50m/s
distance = -83.33m
<h2>
Explanation:</h2>
The velocity of the particle is given by;
v = 20t² - 100t + 50 -------------------(i)
Since acceleration is the time rate of change in velocity, to get the acceleration (a), we find the derivative of equation (i) with respect to t as follows;
a =
= ![\frac{d(20t^{2} - 100t + 50)}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%2820t%5E%7B2%7D%20%20-%20100t%20%2B%2050%29%7D%7Bdt%7D)
a = 40t - 100 ------------------(ii)
Now, when a = 0, let's find the time t by substituting the value of a into equation (ii) as follows;
0 = 40t - 100
=> 40t = 100
=> t = ![\frac{100}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B100%7D%7B40%7D)
=> t = 2.5seconds.
This means that at t = 2.5 seconds, the acceleration of the particle is zero(0)
(a) Now, to get the velocity at this instant (t = 2.5s), substitute the value of t into equation (i) as follows;
v = 20(0)² - 100(0) + 50
v = 0 - 0 + 50
v = 50 m/s
Therefore, the velocity when a is zero is 50m/s
(b) To get the distance (s) travelled at that instant, we integrate equation (i) as follows;
s =
-----------------------(iii)
Where;
a = the time instant = 2.5 seconds
b = the initial time instant = 0
v = 20t² - 100t + 50
Substitute these values into equation (iii) as follows;
s = ![\int\limits^a_b {(20t^{2} -100t + 50)} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%2820t%5E%7B2%7D%20-100t%20%2B%2050%29%7D%20%5C%2C%20dt)
s =
-
+ 50t ![|^{a}_{b}](https://tex.z-dn.net/?f=%7C%5E%7Ba%7D_%7Bb%7D)
Substitute the values of a and b as follows;
s = [
-
+ 50(2.5)] - [
-
+ 50(0)]
s = [
-
+ 50(2.5)] - 0
s = 104.17 - 312.5 + 125
s = -83.33m
Therefore, the distance traveled at that instant is -83.33m