Why is an object that is changing direction said to be accelerating?

Question

My maths teacher explained this to be me by way of analogy: a car driving around a perfectly circular track would be constantly changing its velocity (while the magnitude of the velocity is not changing, the direction is). Because acceleration is the rate of change of velocity, and the object is changing direction, it is said to be accelerating. This strikes me as an odd definition of acceleration, as surely it still equals $\mathrm{0 ms^{-2}}$, even if the object is changing direction. Nevermind, I thought, it's just a definition.

What is strange is that this seeming technicality actually tells us information. Because the object is accelerating, there must have been a resultant force acting on it (since $F=ma$). This is completely baffling to me — yes, the object must have been 'accelerating', but if the magnitude of that acceleration = $0\mathrm{ms^{-2}}$, it seems certain that $F$ = 0 as well. What am I missing?

Answer 1

ACCELEARTION IS A VECTOR QUANTITY!!!

What your Maths Teacher may have meant is that the object changing direction has different acceleration , in the sense that the magnitude of the acceleration is constant (provided no external force) ; it is only changing it's resultant direction .

In Vector form , A Vector A can be written in terms of it rectangular components :

Hence , $A = Acos\theta + Asin\theta $ , Where $\theta $ Is The angle of direction

Since the object is changing direction , at other instant , let the angle made be $\phi$

Then $A' = Acos\phi + A\sin\phi$

Clearly A is not equal to A' Although magnitude of $A = \sqrt{(Acos\theta)^2 +(Asin\theta)^2 } = A$

And magnitude of $A' = \sqrt{(Acos\phi)^2 +(Asin\phi)^2 } = A$

Both acceleration vector have the same magnitude but acts along different direction, so acceleration is changing.

Answer 2

Acceleration means the rate of change of velocity. Here velocity is a vector which can have the same magnitude in different directions. You have considered only the magnitude, not the direction.

$$\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{t}$$

Here $a$ is average acceleration. Consider an object moving in x-y plane along a circular path of radius $1$ centred at the origin with a constant speed $2 m/s$. Velocity of it at $(1,0)$ is tangential to the path and is $2j$. At point $(0,1)$ velocity is tangential to the path and is $-2i$. Time taken by it to travel between these two points is $(distance/speed)$, i.e $π/4$. so average acceleration is:

$$\overrightarrow{a}=\frac{-2i+2j}{\pi/4}\neq0$$

Answer 3

If it is changing direction, then its motion is changing too. This is intuitively what we understand by acceleration. For example, if you were in a vehicle which is changing its direction only, you would feel the changes in the motion of the vehicle.

It's the change in motion (velocity) that we call acceleration.