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...
i really don't want it to be answered i just would love to get help and an explanation on how to answer it

Answer 1

Answer:

$x_{11} = -36 m$

Explanation:

Given

Velocity at t = 0 is $v_{o}$ = 24 m/s

Velocity at t = 1 is also $v_{o}$ = 24 m/s

Acceleration a = -6 $m/s^{2}$

Solution

Duration

$\Delta t = 11 - 1 = 10 s$

Displacement

$S = v_{o} \Delta t + \frac{1}{2} a( \Delta t)^{2}\\\\S = 24 \times 10 + \frac{1}{2} \times (-6) \times (10)^{2}\\\\S = 240 - 300\\\\S = -60 m$

Position at t = 1 s is

$x_{1} = v_{0} t\\\\x_{1} =24 \times 1\\\\x_{1} = 24 m$

Position at t = 11 s is

$x_{11} = x_{1} + S\\\\x_{11} = 24 + (-60)\\\\x_{11} = -36 m$

The body is at 36 meters from the origin in negative x axis

Answer 2

The important thing in this type of questions is to keep track of what you are doing at every point.

From t=0 to t=1s, there is uniform motion at v=24 m/s.

Therefore, it will move 24m in that second of time.

Afterwards, an acceleration comes in, so from t=1s to 11s there will be uniformly accelerated motion with acceleration -6m/s^2.

To account for this, you need to use Suvat's equation:

x(t)=x0+v0*t+1/2*a*t^2.

You should know what to plug in for each of the symbols in this equation:

x0 [initial position at t=1s] = 24m

v0 [initial velocity at t=1s] = 24 m/s

a [acceleration, switched on at t=1s] = -6 m/s^2

t [time from the start of the acceleration until the end, i.e. from t=1s to t=11s] = 10s

Plugging in those numbers in the equation will give you the position at t=11s.