An exercise in Mechanics in As .

Question

an object travels 10 m during one second and 15 m during the next second. find the acceleration of the object assuming it to be constant then find how fast is the object going at the end of the two seconds

Answer 1

Hi Nona!

 

I think you are on the right track with your solution.  There is a deceptive trickiness to this question that I think you picked up on, in that we know neither the acceleration NOR the initial velocity, the latter of which will be very important to determining the final velocity.

 

What I did was to consider information over two intervals that would allow me to make two equations for these two unknowns.  I assume some initial velocity v_o and a constant acceleration a.

 

The first interval is the initial second, over which we also know:

 

time (t) = 1 s

displacement (Δx) = 10 m

 

Then:

 

Δx = v_ot + (1/2)at²

 

10 m = v_o(1 s) + (1/2)a(1 s)² --> 10 = v_o+ a/2

 

Now, for the other equation, I look at the full interval over the first two seconds.  I start with the same v_o, and have the same acceleration a.  This time:

 

t = 2 s

Δx = 25 m

 

So:

 

25 m = v_o(2 s)+(1/2)a(2 s)² -->  25 = 2v_o + 2a

 

Now we have a system of two equations with two unknowns, v_o and a.  Eventually, we need to solve for both, but let's start with a.  So the first equation becomes:

 

v_o = 10 - a/2  (just an algebraic rearrangement of the first expression above)

 

Then substitute this into the second equation, and solve for a:

 

25 = 2(10 - a/2) + 2a 

25 = 20 - a + 2a

a = 5 m/s²

 

Both you and Adam got the same answer.  Why is this?  What Adam calculated over the first two seconds was actually the average velocity over each second.  But, if acceleration is constant over an interval, it turns out that the average velocity over that interval equals the instantaneous velocity at the time midpoint of the interval.  So, in the first interval, the velocity is EXACTLY 10 m/s at t = 0.5 s.  And the velocity is exactly 15 m/s at t = 1.5 s in the second interval.  The definition for acceleration involves the difference of instantaneous velocities, so you can then write:

 

a = (15 m/s - 10 m/s)/(1.5 s - 0.5 s) = 5/1 = 5 m/s²

 

However, Adam's solution assumed the initial velocity was zero.  And, if you start from rest with a 5 m/s² acceleration, you do not cover 10 m in the first second or 15 m in the second second.  This is because there is some non-zero initial velocity (I think Adam also accidentally solved for displacement instead of velocity).  We can now calculate v_o from the system of equations above, because:

 

v_o = 10 - a/2 = 10 - 5/2 = 7.5 m/s

 

Starting at this velocity does give you 10 m over the first second and 15 m over the second second with a 5 m/s² acceleration.

 

We can now determine the velocity (v) at the end of the two seconds using another kinematic equation:

 

v = v_o+at = 7.5 m/s + (5 m/s²)(2 s) = 17.5 m/s

 

I hope this helps!  If you have any questions about any of this, please let me know.

 

 

Answer 2

 

Let v1 = the velocity at 1 second

From the problem, we know that v1 = 10m/s 

 

Let v2 = the velocity at 2 seconds

From the problem, we know that v2 = 15m/s

 

We know the acceleration is constant (from the problem), so we can figure out the acceleration by comparing the velocity at the two different times:

 

We know that acceleration = change in velocity / change in time

So we have:

 

a = (15 - 10) / (2 - 1)

a = 5/1 = 5m/s²

 

Therefore, the acceleration is 5m/s2.

 

Therefore, the total velocity at the end of two seconds is 20m/s