Question

) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].

Answer 1

The displacement (net distance covered) of the particle during the time interval [-2,5] is -29/6.

Step-by-step explanation:

The velocity function is given as v(t) = -t2 + 3t - 2.

To find the displacement of the particle during the time interval [-2,5], we need to integrate the velocity function over that interval.

Integrating v(t) over [-2,5] gives us the displacement D:

D = ∫(-t2 + 3t - 2) dt = [-t3/3 + (3t2)/2 - 2t] evaluated from -2 to 5

Simplifying, we find D = (-125/3 + 75/2 - 10) - (-8/3 + 6 - 4) = -29/6

So, the displacement covered by the particle during the time interval [-2,5] is -29/6.

Answer 2

89.87m/s

Step-by-step explanation:

Given the velocity function

v(t)=−t²+3t−2

In order to get the displacement function, we will integrate the velocity function as shown:

`\int\limits^5_(-2) {v(t)} \, dt \\d(t)= \int\limits^5_(-2){(-t^2+3t+2)} \, dt \\\\d(t)=[(-t^3)/(3)+(3t^2)/(2)+2t ]^5_(-2)\\`

at t = 5

`d(5)=[(-5^3)/(3)+(3(5)^2)/(2)+2(5) ]\\d(5)=[(-125)/(3)+(75)/(2)+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s`

at t = -2

`d(-2)=[(-(-2)^3)/(3)+(3(-2)^2)/(2)+2(-2) ]\\d(-2)=[(-8)/(3)+(12)/(2)+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s`

Required displacement = d(5) - d(-2)

Required displacement = 89.2 - (-0.67)

Required displacement = 89.2 + 0.67

Required displacement = 89.87m/s