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 position x of the body at 11 seconds is "-36 m".

According to the question,

Distance covered in 1 second,

• = $24\times 1$

        = $24 \ m$

As we know,

→ $s = ut+0.5\times at^2$

Distance covered in at 10 more seconds,

→    $= 24\times 10-0.5\times 6\times 10^2$

→    $=240-0.5\times 6\times 100$

→    $= 240-300$

→    $= 60 \ m$

hence,

The total distance covered will be:

= $24-60$

= $-36 \ m$  

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