Question

The velocity of a particle is given by v=20t² - 100t + 50, where v is in meters per second and t is in seconds. Evaluate the velocity when a is zero and the distance travelled at that instant.

Answer 1

velocity = 50m/s

distance = -83.33m

Step-by-step explanation:

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 =
`(dv)/(dt)` =
`(d(20t^(2) - 100t + 50))/(dt)`

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 =
`(100)/(40)`

=> 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 =
`\int\limits^a_b {v} \, dt` -----------------------(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`

s =
`(20t^(3) )/(3)` -
`(100t^(2) )/(2)` + 50t
`|^(a)_(b)`

Substitute the values of a and b as follows;

s = [
`(20(2.5)^(3) )/(3)` -
`(100(2.5)^(2) )/(2)` + 50(2.5)] - [
`(20(0)^(3) )/(3)` -
`(100(0)^(2) )/(2)` + 50(0)]

s = [
`(20(2.5)^(3) )/(3)` -
`(100(2.5)^(2) )/(2)` + 50(2.5)] - 0

s = 104.17 - 312.5 + 125

s = -83.33m

Therefore, the distance traveled at that instant is -83.33m

Answer 2

Step-by-step explanation:

It is given that,

The velocity of a particle is given by :

`v=20t^2-100t+50`

Where

v is in m/s and t is in seconds

Let a is the acceleration of the object at time t. So,

`a=(dv)/(dt)`

`a=(d(20t^2-100t+50))/(dt)`

`a=40t-100`

When a = 0

`40t-100=0`

t = 2.5 s

a is zero at t = 2.5 s. Velocity,
`v=20(2.5)^2-100(2.5)+50`

v = -75 m/s

Since,
`v=(ds)/(dt)`, s is the distance travelled

`s=\int\limits{vdt}`

`s=\int\limits{(20t^2-100t+50)dt}`

`s=(20t^3)/(3)-50t^2+50t`

At t = 2.5 s,
`s=(20(2.5)^3)/(3)-50(2.5)^2+50(2.5)`

s = −83.34 m

Hence, this is the required solution.