Question

A point mass moves with velocity v=(5t-1²) ms' in a straight line. Find the distance travelled (i.e.
in fourth second.
31
(B)
m
(D) None of these​

Answer 1

Given: The correct equation is v = 5t - t^2

To find: the distance in fourth second?

Solution:

• Now we have given the equation as:

                 v = 5t - t^2

• It can be written as:

                 dx/dt = 5t - t^2

• Integrating both sides, with lower limit 3 and upper limit 4, we get:

                 x =   $\int\limits^4_3 {5t - t^2} \, dt$

                 x = (5t^2/2 - t^3/3)

• Putting the limits, we get:

                 x = (5(4)^2/2 - (4)^3/3) - (5(3)^2/2 - (3)^3/3)

                 x = (5x16/2 - 64/3) - (5x9/2 - 27/3)

                 x = (40 - 64/3) - (45/2 - 9)

                 x = (120 - 64 / 3) - (45 - 18 / 2)

                 x = 56/3 - 27/2

                 x = 112 - 81 / 6

                 x = 31 / 6

Answer:

             So the distance travelled in fourth second is 31/6 m