Question

The velocity of a particle which moves along the s-axis is given by v = 2-4t+5t^(3/2), where t is in seconds and v is in meters per second. Evaluate the position s, velocity v, and acceleration a when t=4s. The particle is at the position s0 =2m when t=0

Answer 1

Answer:

The position s, velocity v, and acceleration a when t = 4s are 40m, 26m/s and 11m/s² respectively.

Step-by-step explanation:

The velocity, v, of the particle at any time, t, is given by;

v = 2 - 4t + 5
`t^{(3)/(2) }` ----------(i)

Analysis 1: To get the position, s, of the particle at any time, t, we integrate equation (i) with respect to t as follows;

s = ∫ v dt

Substitute the value of v into the above as follows;

s = ∫ (2 - 4t + 5
`t^{(3)/(2) }`) dt

s = 2t -
`(4t^2)/(2)` +
`(5t^(5/2))/(5/2)`

s = 2t -2t² + 2
`t^{(5)/(2) }` + c [c is the constant of integration] ------------(ii)

According to the question;

when t = 0, s = 2m

Substitute these values into equation (ii) as follows;

2 = 2(0) -2(0)² + 2
`(0)^{(5)/(2) }` + c

2 = 0 - 0 - 0 + c

c = 2

Substitute the value of c = 2 back into equation (ii) as follows;

s = 2t -2t² + 2
`t^{(5)/(2) }` + 2 --------------------------------(iii)

Analysis 2: To get the acceleration, a, of the particle at any time, t, we differentiate equation (i) with respect to t as follows;

a =
`(dv)/(dt)`

Substitute the value of v into the above as follows;

a =
`(d(2 - 4t + 5t^(3/2)))/(dt)`

a = -4 +
`(15)/(2)`
`t^(1/2)` ----------------------------------(iv)

Now;

(a) When t = 4s, the position s, of the particle is calculated by substituting t=4 into equation (iii) as follows;

s = 2(4) -2(4)² + 2(
`4^{(5)/(2) }`) + 2

s = 8 - 32 + 2(4)²°⁵

s = 8 - 32 + 2(32)

s = 8 - 32 + 64

s = 40m

(b) When t = 4s, the velocity v, of the particle is calculated by substituting t=4 into equation (i) as follows;

v = 2 - 4(4) + 5(
`4^{(3)/(2) }`)

v = 2 - 16 + 5(4)¹°⁵

v = 2 - 16 + 5(8)

v = 2 - 16 + 40

v = 26m/s

(c) When t = 4s, the acceleration a, of the particle is calculated by substituting t = 4 into equation (iv) as follows;

a = -4 +
`(15)/(2)`(
`4^(1/2)`)

a = -4 +
`(15)/(2)`(2)

a = -4 + 15

a = 11m/s²

Therefore, the position s, velocity v, and acceleration a when t=4s are 40m, 26m/s and 11m/s² respectively.

Answer 2

a) s = 42 m

b) V = 26 m/s

c) a = 11 m/s²

Step-by-step explanation:

The velocity(V) = 2 - 4t + 5t^3/2 where t is in seconds and V is in m/s

a) To get the position, we have to integrate the velocity with respect to time.

We get the position(s) from the equation:

`V=(ds)/(dt)`

`ds=Vdt`

Integrating both sides,

`s=\int\ {V} \, dt`

`s=\int\ {(2-4t+5t^{(3)/(2) }) \, dt`

Integrating, we get

`s=(2t-2t^(2) +2t^{(5)/(2) }+c)m`

But at t=0, s(0) = 2 m

Therefore,

`s(0)=2(0)-2(0)^(2) +2(0)^{(5)/(2) }+c`

`2=0+0+0+c`

c = 2

Therefore

`s=(2t-2t^(2) +(10)/(5)t^{(5)/(2) }+2)m`

When t = 4,

`s=2(4)-2(4)^(2) +2(4)^{(5)/(2) }+2`

`s=8-32+64+2`

s = 42 m

At t= 4s, the position s = 42m

b) The velocity equation is given by:

`V=(2-4t+5t^{(3)/(2) })m/s`

At t = 4,

`V=2-4(4)+5(4)^{(3)/(2) }`

V = 2 - 16 + 40 = 26 m/s

The velocity(V)at t = 4 s is 26 m/s

c) The acceleration(a) is given by:

`a=(dv)/(dt)`

`a=(d)/(dt)(2-4t+5t^{ (3)/(2)})`

Differentiating with respect to t,

`a=-4+(15)/(2)t^{(1)/(2) }`

`a=(-4+(15)/(2)t^{(1)/(2) })m/s^(2)`

At t = 4 s,

`a=-4+(15)/(2)(4)^{(1)/(2) }`

`a=-4+15`

a = 11 m/s²