2
$\begingroup$

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

My attempt at solving the problem: $$a(x) = \frac{v - u}{t}$$

where

$v =$ final velocity

$u =$ initial velocity $$$$

I get the answer as $4.05 \space ms^{-2}$

But the correct answer given to the problem is $8.10 \space ms^{-2}$.

They used a different equation to reach that answer.

Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity?

$\endgroup$
  • 5
    $\begingroup$ Your procedure is not wrong, but which values for $v$ and $u$ have you put in? They don't appear in the question. You must be doing another step as well to find the velocities. $\endgroup$ – Steeven Feb 9 '17 at 9:59
  • $\begingroup$ If you solved a problem, got wrong answer, and want to ask for help, it would help if you actually show how you solved the problem. "I used this equation and got this answer" isn't usually enough to help. Here, your error most likely is that you used a correct equation but in the wrong way, but there's no way we can know what you did wrong, although many might try to guess if they have seen similar mistakes by themselves other people. $\endgroup$ – JiK Feb 9 '17 at 10:48
  • $\begingroup$ BTW, describing the acceleration of a car as uniform is one of the cases where spherical cow reasoning is misplaced. The acceleration of a car can only very badly approximated as uniform, because the force is nowhere near constant; rather, the power is constant, which means the acceleration decreases at speed. (Though for a modern Tesla in Ludicrous Mode it might actually be appropriate, because the traction control is able to regulate almost constant force – just enough so the wheels won't spin.) $\endgroup$ – leftaroundabout Feb 9 '17 at 14:26
5
$\begingroup$

That equation is not the most straight foreward for this situation but it can be used, here's how. If you divide the displacement, 110 m, by the time of 5.21 s you will get 21.1 m/s. But that is the average velocity over the entire displacement. The average veloctiy, when v is changing uniformly, is found by adding the initial and final velocities and dividing by two. Since the initial velocity is 0 m/s, you will need to double the 21.1 m/s for the final velocity such that the average will be the calculated 21.1 m/s. So the final velocity, and therefore the change in velocity over that displacement, is 42.2 m/s. When you divide that delta v by the time over which it occurs, you will have how much delta v per second, acceleration.

While it's true there is another equation that does this all in one step, it would be good for you to understand the process I've outlined above.

$\endgroup$
  • $\begingroup$ And you know v is changing uniformly, because acceleration is changing uniformly, correct? $\endgroup$ – Nak Leng Feb 9 '17 at 14:16
  • $\begingroup$ @NakLeng No, acceleration is not changing, it is just constant. $\endgroup$ – Arvo Feb 9 '17 at 14:51
  • $\begingroup$ @NakLeng. The acceleration is uniform but not changing, it is constant. At this level of Phyics, you will not be subjected to numerical problems with changing acceleration (jerk). The result of constant acceleration is a straight line v-t plot. Look at Farcher's answer, his sketch shows my description. $\endgroup$ – bpedit Feb 9 '17 at 14:56
4
$\begingroup$

What is your $V_f$ ?

$V_f$ is not given in the question so you can't use this equation, $$V_f =V_i +at$$

But the distance is given which will allow you to use $$S=ut+1/2 at^2 $$

$\endgroup$
  • $\begingroup$ Well, they gave me meters and they gave me seconds, so I divided the two to get 21.1 m/s as final velocity. Am I wrong in assuming that this would be the final velocity? $\endgroup$ – Nak Leng Feb 9 '17 at 3:04
  • 4
    $\begingroup$ Yes,you are wrong ,thats not the final velocity $\endgroup$ – Lapmid Feb 9 '17 at 3:06
  • 1
    $\begingroup$ @NakLeng Of course :) What you actually calculated was average velocity. Think about it - the car was moving slower at the start, so under linear acceleration you'd need to "catch up" to the relatively slower first half by moving faster in the second. Going from a stand, this means the final velocity is twice the average (depending on your skill level, use either geometry or integral count to prove this) - which is why your answer was half the acceleration of the correct answer. $\endgroup$ – Luaan Feb 9 '17 at 12:03
2
$\begingroup$

The displacement is equal to the area under a velocity $v$ against time $t$ graph as shown below.

enter image description here

If the body starts from rest and its final velocity is $v_{\rm f}$ then the average velocity is $\dfrac{v_{\rm f}}{2}$ and that is were your missing "$2$" comes from.

$\endgroup$
1
$\begingroup$

Yes, I think so. Below is the proper formula for the distance an object accelerating at a constant rate goes over time. I used it to get a formula where you enter distance and time traveled to get the acceleration. m=0.5at^2 2m/t^2=a (2*110)/5.21^2=8.10489203915 m/s^2

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.