Q. An object starts at time t = 0 with a velocity of v0 = + 21 m/s and undergoes a constant acceleration of a = -5 m/s^2.(adsbygoogle = window.adsbygoogle || []).push({});

a) How far from its starting ( t = 0 ) is the object at time at which it comes to rest?

Solution:

Since, the object undergoes constant acc. constant acceleration equations may be applied.

Therefore,

We need to find the time 't' first,

so, v = v0 + at ( It is going to rest , so v = 0 m/s )

-v0 = at

Given a = -5 m/s^2.

Plugging in we get t = 4.2 seconds, which I believe is right.

Then, using:

x -x0 = v0t - 1/2 a*(t2-t1)^2

x0 ----> 0

v0-----> +21 m/s

't' from above ------> 4.2 seconds which is 't2) and t1 = 0 ---> at rest.

'a' from above -------> -5 m/s^2

Plugging in all the know data in the above eqn, then give us.

x = (21 m/s)*(4.2 seconds) - 1/2 ( -5 m/s^2) * ( t2-t1)^2

I got x = 88.2 + 44.1 = 132.2 meters.

Is this correct!

b) Suppose a second object begins moving with a constant speed of v = 19 m/s in the same direction from the same location 3 seconds later. At what time do the paths of these two objects again cross?

If the objects cross, then their x - coordinates, may be set equal to one another and solved for time. That's all I know. I tried to setup by doing the following:

1st case:

v0 = +21 m/s

a = -5 m/s^2

t = 0 at rest

2nd case :

v0 = 19 m/s

and since its constant speed acc. - 0.

setting up the constant acc,equations for the 1st case data:

x = (21m/s)t - 1/2 (-5 m/s^2)t^2

setting up the constant acc, equations for the 2nd case data:

x = (19m/s)t [ and the rest of the terms vanish since speed is constant for the second case )

equating the above we get:

21t + 2.5t^2 = 19t

then, becomes quadratic.

2.5t^2-19t+21t = 0

is this correct so far. Is my reasoning correct for both the parts so far.

and my work!

Can one be kind enough to tell me if something is wrong.

Thanks, for looking

and appreciate some feedback!

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# Homework Help: 1-D Motion

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