Question

An object moves along a straight line. The position as a function of time is given by the following
22 formula x(t)=5m+(2m/s)t–(0.6m/s)t^2. Please answer all of the parts, I will give a good rating**** :)

How long after starting does it take the object to pass the origin?
At what time is the object located 4 m from the origin?
Sketch the following graphs: x(t), v(t), a(t).
Find the object’s initial acceleration.
At what time will the object come to rest?

Answer 1

Given the expression for the position as a function of time

where, distance is in meter and time is in second, and the above parameters are
  
And so, without writing the units explicitly, we write

So, when the object passes the origin, we have x(t) = 0, i.e.,
  
Solving this quadratic equation for t, we get

  
  

But as time can't be negative, so, we take only the positive root. And so, the required time to cross the origin is
  

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The time at which the object is located at 4m from the origin (towards the positive x direction ) is given by

And solving this quadratic equation again as before and taking only the positive root, we get the required time
  
However if the object is located towards the negative x direction at a distance of 4m from the origin, then the required time to reach that position is computed as follows

And solving this quadratic equation again as before and taking only the positive root, we get the required time
  
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We now write the position x(t), velocity v(t) and the acceleration a(t) as follows :
Given
  
the velocity as a function of time is

And the acceleration as a function of time is
  
The acceleration is independent of time. The sketches are

And
  
And
  

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As we have obtained already that

, and it is time independent,
so, the initial acceleration of the object is

And the negative sign implies that the object is actually decelerating.
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The object will come to rest at a time when the velocity of the object will be zero, i.e.,