Question

A body moving in the positive x direction passes the origin at time t = 0. between t = 0 and t = 1 second, the body has a constant speed of 24 meters per second. at t = 1 second, the body is given a constant acceleration of 6 meters per second squared in the negative x direction. the position x of the body at t = 11 seconds is

Answer 1

Integrate over [0, 1] to get the position of the body after 1 second. It starts at the origin, so we take x(0 s) = 0 m and using the fundamental theorem of calculus,

`\displaystyle x(1\,\mathrm s) = x(0\,\mathrm s) + \int_(0\,\rm s)^(1\,\rm s) \left(24(\rm m)/(\rm s)\right) \, dt = 24 \, \mathrm m`

Then for time t ≥ 1 second, we integrate over [1, 11] to get the position after the next 10 seconds of motion. Since x(1 s) = 24 m, we have

`\displaystyle x(11\,\mathrm s) = x(1\,\mathrm s) + \int_(1\,\rm s)^(11\,\rm s) \left(24(\rm m)/(\rm s) - \left(6(\rm m)/(\mathrm s^2)\right) t\right) \, dt = \boxed{-36\,\mathrm m}`

Without using calculus, in the 1st second of motion the body travels at constant speed, so it reaches a new position of

x = (24 m/s) (1 s) = 24 m

In the next 10 seconds, the body has constant acceleration and its velocity at time t is

v = 24 m/s - (6 m/s²) (t - 1 s)

so its "initial" velocity is 24 m/s, and after 10 seconds it attains a "final" velocity of

24 m/s - (6 m/s²) (11 s - 1 s) = -36 m/s

Then the body travels a distance x such that

(-36 m/s)² - (24 m/s)² = 2 (-6 m/s²) x ⇒ x = - 60 m

and so the body's net change in position is 24 m - 60 m = -36 m.