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Acceleration is change of directions as well, let’s assume an object is travelling at 10 mph north, and turned west by the same speed of 10 mph to west and kept going that way. What would be the acceleration? As this is L shaped movement, I don’t think we could use angular acceleration formula to find the acerbation, how could you address this?

Abiy Ermias
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Acceleration is the rate of change of velocity, so all we have to do to find the acceleration is find the change in velocity, $\Delta v$, and divide it by the time taken for the velocity to change, $\Delta t$. More precisely this gives us the average acceleration. For the instantaneous acceleration we need to take the limit of $\Delta t \to 0$.

Anyhow suppose in your example the car takes ten seconds to turn the corner and to keep things metric take its speed to be $10$ m/s. If we take East to be the $\mathbf{i}$ direction and North to be the $\mathbf{j}$ direction then the initial velocity is:

$$ \mathbf{u} = 0\mathbf{i} + 10\mathbf{j}~\textrm{m/s} $$

and the final velocity is:

$$ \mathbf{v} = -10\mathbf{i} + 0\mathbf{j}~\textrm{m/s} $$

Then the change in velocity is the vector:

$$ \Delta \mathbf{v} = \mathbf{v} - \mathbf{u} = -10\mathbf{i} - 10\mathbf{j}~\textrm{m/s} $$

Then divide by the time, $\Delta t = 10$ secs to get the average acceleration:

$$ a = -1\mathbf{i} - 1\mathbf{j}~\textrm{m/s}^2 $$

The acceleration is a vector instead of just a number, as we get in 1D calculations, but that's what we would expect since as you say direction matters here. If you want a single number so you can calculate the g-force on the driver just take the magnitude of the vector:

$$ |\mathbf{a}| = \sqrt{2}~\textrm{m/s}^2 $$

John Rennie
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  • "Anyhow suppose in your example the car takes ten seconds to turn the corner". It seems to me the problem with the OP example is the object is not "turning a corner" if its movement is "L" shaped, which would imply a zero radius turn requiring an infinite centripetal acceleration. – Bob D Oct 31 '23 at 14:47
  • @BobD Let's make the effort to be helpful if we can. – John Rennie Oct 31 '23 at 14:49
  • Doesn't an L shaped movement imply $\Delta t=0$? – Bob D Oct 31 '23 at 14:50
  • I suspect you are taking the OP's way of phrasing things excessively literally. – John Rennie Oct 31 '23 at 14:51
  • I wasn't criticizing your answer and I do want to be helpful. I'm just concerned that the last sentence in the OP suggests he is literally thinking about an L movement. Anyway, let's just leave it there. – Bob D Oct 31 '23 at 15:00
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As has been pointed out there can be no right-angled bend in which the objects moves at constant speed but instantaneously changing direct as this would be an infinite acceleration.

However one could consider the right-angled bend to be quadrant of a circle of radius $r$.
Assuming that the peed of the object stayed at $5\,\rm m/s (\approx 10\,mph)$ then one could say that the centripetal acceleration of the object was $10^2/r= 100/r$.

So now all that is required is a value of the radius of the quadrant and to note that the magnitude of the acceleration stays constant but its direction starts in a from Westerly direction, changing smoothly to a Southerly direction.

Farcher
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An object cannot instantaneously change its direction of travel by 90 degree. But an approximation of such an 'L'-shaped motion is a billiard ball cannoning off a cushion at a 45 degree incidence angle. In that case, a reasonable approximation of the acceleration resulting from the collision of a north-going ball that bounces off in a west-going direction would be an 'infinite' acceleration for a 'zero' time interval, where the acceleration is in the south west direction.

In practice, that is impossible, so there wil be a finite acceleration over a finite period of time - with the acceleration still (approximately) in the south-west direction. You will get a better understanding of what is happenig during the change in direction by looking up the term impulse in a text-book covering kinematics (or a physics website).

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Penguino
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But as long as the magnitude of velocity doesnt change...acceleration would be zero no mattet what the direction is..i mean there would be a acceleration vector which would be a null vector..and null vectors direction is undefined

Pheonix
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  • Centripetal acceleration always points perpendicular to the velocity, and changes only the direction but not the magnitude of the velocity. That doesn't mean the acceleration is zero. A constant-magnitude velocity that's changing direction is not a constant velocity, and does not imply zero acceleration. – Nuclear Hoagie Oct 31 '23 at 14:47