Mechanics problem — Car moving with constant acceleration

[Vortex8380]

Homework Statement

A car is moving at a constant acceleration on a horizontal road passing points A, B and C. The car travels from A to B (22 metres) in 2 seconds and then passes B to C (104 metres) in 4 seconds. What is the acceleration of the car?

Assume that acceleration from A to C is constant

Relevant Equations

SUVAT Equations

I tried this but I don't know if it makes sense:

Average velocity from A to B = 22/2 = 11m/s

Average velocity from B to C = 104/4 = 26m/s

(26-11)/6 = 3.75m/s

 

[jbriggs444]

I tried this but I don't know if it makes sense:

Average velocity from A to B = 22/2 = 11m/s

Average velocity from B to C = 104/4 = 26m/s

(26-11)/6 = 3.75m/s

At what time is the instantaneous velocity of the car equal to the 11 m/s average? Is it at the beginning of the interval?

At what time is the instantaneous velocity of the car equal to the 26 m/s average? Is it at the end of the interval?

Edit: +1 to the suggestion by @PeroK It is always wise to sanity check your results.

 

[PeroK]

Homework Statement:: A car is moving at a constant acceleration on a horizontal road passing points A, B and C. The car travels from A to B (22 metres) in 2 seconds and then passes B to C (104 metres) in 4 seconds. What is the acceleration of the car?

Assume that acceleration from A to C is constant
Homework Equations:: SUVAT Equations

I tried this but I don't know if it makes sense:

Average velocity from A to B = 22/2 = 11m/s

Average velocity from B to C = 104/4 = 26m/s

(26-11)/6 = 3.75m/s

That was a good try. You could have tested your answer. If $a = 3.75 m/s^2$, then what was its velocity at points A, B and C?

See also the post above!

 

[Vortex8380]

Im not sure, i just took the average velocities of between the 2 distances and asssuming i could use them as the inital and final velocities over the total amount of time taken to get from A to C

 

[PeroK]

Im not sure, i just took the average velocities of between the 2 distances and asssuming i could use them as the inital and final velocities over the total amount of time taken to get from A to C

If $a = 3.75 m/s^2$, then $v_B = v_A + 7m/s$ and the average velocity between A and B is $v_a + 3.75m/s$.

We know that this should be $11m/s$, which gives us $v_A = 7.25 m/s$.

Does it all work out between B and C with $v_A = 7.25 m/s$ and $a = 3.75 m/s^2$?

 

[jbriggs444]

Im not sure...
i just took the average velocities of between the 2 distances and asssuming i could use them as the inital and final velocities over the total amount of time taken to get from A to C

OK. Let us make the hint more obvious.

If you are computing acceleration by taking the difference between final and initial velocities and dividing by the time interval then you must be assuming that the initial velocity, exactly at time zero is 11 m/s and that the final velocity exactly at time 6 is 26 m/s.

Is the initial velocity equal to 11 m/s? Well, let us perform a sanity check...

If the initial velocity is 11 m/s and the object accelerates at 3.75 m/s² then the velocity after 2 seconds will be somewhat higher. And the distance traveled in those two seconds will be somewhat greater than 2 s * 11 m/s = 22 meters.

But the problem statement says that the distance traveled in those two seconds was 22 meters. So that 11 m/s cannot be correct.

Edit: More directly, if the object's average velocity over the first two seconds is equal to its initial velocity at the beginning of those two seconds then its acceleration must be zero. But that obviously does not fit the problem statement.

 

[Vortex8380]

ok thank you

OK. Let us make the hint more obvious.

If you are computing acceleration by taking the difference between final and initial velocities and dividing by the time interval then you must be assuming that the initial velocity, exactly at time zero is 11 m/s and that the final velocity exactly at time 6 is 26 m/s.

Is the initial velocity equal to 11 m/s? Well, let us perform a sanity check...

If the initial velocity is 11 m/s and the object accelerates at 3.75 m/s² then the velocity after 2 seconds will be somewhat higher. And the distance traveled in those two seconds will be somewhat greater than 2 s * 11 m/s = 22 meters.

But the problem statement says that the distance traveled in those two seconds was 22 meters. So that 11 m/s cannot be correct.

ok thank you

 

[archaic]

You have this fact for 2nd degree polynomials: $\frac{f(b)-f(a)}{b-a}=f'\left(\frac{b+a}{2}\right)$. In your case, look at the function $f'$ as the velocity, and $f$ as the position (which is the second degree polynomial, because the acceleration is constant).
For the sake of explanation, if the position is given by $x(t)$, then $\frac{x(t_2)-x(t_1)}{t_2-t_1}=\frac{\Delta x}{\Delta t}$, is the average velocity over the period $\Delta t$, but it also is equal to the instantaneous velocity at $t=$ half of the time between the start and the end.

 

[jbriggs444]

(26-11)/6 = 3.75m/s

Missed this on the first read-through

$\frac{26-11}{6}$ is not equal to 3.75. $\frac{26-11}{4}$ is equal to 3.75. Why did you divide by four when saying you were dividing by six?