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:

a) s = 42 m

b) V = 26 m/s

c) a = 11 m/s²

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:

$V=\frac{ds}{dt}$

$ds=Vdt$

Integrating both sides,

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

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

Integrating, we get

$s=(2t-2t^{2} +2t^{\frac{5}{2} }+c)m$

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

Therefore,

$s(0)=2(0)-2(0)^{2} +2(0)^{\frac{5}{2} }+c$  

$2=0+0+0+c$

c = 2

Therefore

$s=(2t-2t^{2} +\frac{10}{5}t^{\frac{5}{2} }+2)m$

When t = 4,

$s=2(4)-2(4)^{2} +2(4)^{\frac{5}{2} }+2$

$s=8-32+64+2$

s = 42 m

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

b) The velocity equation is given by:

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

At t = 4,

$V=2-4(4)+5(4)^{\frac{3}{2} }$

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:

$a=\frac{dv}{dt}$

$a=\frac{d}{dt}(2-4t+5t^{ \frac{3}{2}})$

Differentiating with respect to t,

$a=-4+\frac{15}{2}t^{\frac{1}{2} }$

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

At t = 4 s,

$a=-4+\frac{15}{2}(4)^{\frac{1}{2} }$

$a=-4+15$

a = 11 m/s²

Answer 2

Answer:

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

Explanation:

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

v = 2 - 4t + 5$t^{\frac{3}{2} }$          ----------(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$t^{\frac{3}{2} }$) dt          

s = 2t - $\frac{4t^2}{2}$ + $\frac{5t^{5/2}}{5/2}$

s = 2t -2t² + 2$t^{\frac{5}{2} }$ + 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$(0)^{\frac{5}{2} }$ + c  

2 = 0 - 0 - 0 + c

c = 2

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

s = 2t -2t² + 2$t^{\frac{5}{2} }$ + 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 = $\frac{dv}{dt}$

Substitute the value of v into the above as follows;

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

a = -4 + $\frac{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($4^{\frac{5}{2} }$) + 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($4^{\frac{3}{2} }$)

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 + $\frac{15}{2}$($4^{1/2}$)

a = -4 + $\frac{15}{2}$(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.