Question

Given v(t) = 25+18t, where v is in m/s and t is in s, use calculus to
determine the total displacement from t1=1.5s to t2 =3.5 s ∆r=__m​

Answer 1

Answer:

v= dr/dt = 25+18t

integrating v from 1.5 to3.5

r = 25t+ 18t^2 /2

r = (25(3.5)+ 9(3.5)^2) - (25(1.5)+9(1.5))

Answer 2

Answer:

The correct answer is  106 m

Explanation:

From the above question,

They have given :

υ(t) = 25 + 18t,

where υ is in m/s and t is in s, use calculus to determine the total displacement  

from t1 = 1.5 s to t2 = 3.1 s.

To find the total displacement between t1 = 1.5 s and t2 = 3.5 s:

r(t) = ∫ v(t) dt

r(t) = ∫ (25 + 18t) dt

r(t) = 25t + $\frac{9t^2}{2}$ + C (where C is the constant of integration)

To find the constant of integration C:

                   r(t1) = 25t1 +  $\frac{9t^2}{2}$ + C = 0

                   C = -25t1 -  $\frac{9t^2}{2}$

Substituting the value of C, we get:

                   r(t) = 25t +  $\frac{9t^2}{2}$  - 25t1 -  $\frac{9t^2}{2}$ +

Now we can calculate the total displacement ∆r between t1 = 1.5 s and t2 = 3.5 s by subtracting the displacement at t1 from the displacement at t2:

                   ∆r = r(t2) - r(t1)

∆r = $[25(3.5) + \frac{9(3.5)^2}{9} \frac{(3.5)^2}{2} - 25(1.5) - \frac{9(1.5)^2}{2} ] - [25(1.5) + \frac{9(1.5)^2}{2} - 25(1.5) - \frac{9(1.5)^2}{2} ]$

                   ∆r = [126.875] - [33.75]

                   ∆r = 106 meters (rounded to three decimal places)

Therefore, the total displacement between t1 = 1.5 s and t2 = 3.5 s is 106 meters.

The correct answer is  106 m

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