# Direction of acceleration vector at the apex of a curve

(Test Your Understanding of Section 3.2)

The question wants me to choose the correct acceleration vector, from the given 9 options where 9th option is zero acceleration.

Ok, first I thought option 9 (zero acceleration) is the correct answer here. The sled’s speed decreases as it goes up the hill, because in this case there is a parallel component of gravitational acceleration, i.e $$g\sin\theta$$, that is acting opposite to the velocity of the sled. (There is a perpendicular component $$g\cos\theta$$ too, but that is only changing the direction of the velocity).

Likewise, sled’s velocity increases as it goes down the hill because ‘$$g\sin\theta$$’ in this case is acting along the direction of the sled’s instantaneous velocity. But when the sled is at the crest of the curved trajectory, the sled’s velocity instantaneously becomes constant because parallel component of acceleration, i.e $$g\sin\theta$$ is zero. And I thought that the perpendicular component of $$g$$, or $$g\cos\theta$$ is getting cancelled out by the normal force acting on the sled. That’s why I thought option 9 was correct, i.e zero acceleration when the sled is at the crest of the hill.

But the answer at the end of the chapter says that option 7 is correct. How can option 7 be correct, because the sled isn’t moving vertically upwards, or vertically downwards when it is at the crest of the hill. Then how can we say that option 7 is correct?

Also, in the question above that ($$conceptual$$ $$example$$ $$3.4$$), how is the direction of acceleration at $$Point$$ $$E$$ vertically upwards?

You are failing into the trap that many students fall into. Moving in a certain direction does not mean acceleration is in that direction.

Acceleration is a vector quantity that describes how the velocity vector (not the speed) changes. Therefore, there is nothing saying that the direction of motion must coincide with the acceleration (it usually does not). What does point in the direction of motion is the velocity vector, as it always points tangent to the trajectory.

In this case, we are told the speed is increasing up the curve and decreasing down the curve, so you are correct in thinking at at the Apex the speed is not changing. However, the direction of the velocity is still changing, and hence there is still an acceleration. If speed is not changing but the direction of velocity is changing then acceleration must act perpendicular to the path. This is why the answer is what you say it is.

Also be careful about equating the normal force to the weight at the apex. This is not true. $$N=mg\cos\theta$$ is only true if the acceleration is parallel to the path (i.e. no direction change. Just a change in speed). This is not the case here.

• Thanks. I think I understand it now. Speed of the sled is not changing at the apex of the curve, but its direction is changing, that makes sense. Also, now I understand that I was incorrect in thinking that acceleration at the apex was zero. Because if there was zero acceleration, the sled would have continued moving along straight without changing its direction. Thanks for explaining it. – π times e Oct 15 at 7:30
• And how is the normal force getting balanced in this case? Something else must cancel it out because the sled doesn’t move vertically upwards? – π times e Oct 15 at 7:37
• @πtimese As explained in my answer, direction of motion $\neq$ direction of acceleration. In this case the normal force is actually less than the weight of the sled at the apex, hence the net downward acceleration of the sled at this point. – Aaron Stevens Oct 15 at 7:39
• Does that mean the net downward acceleration at that point is less than the magnitude of $g$ ? – π times e Oct 15 at 7:45
• @πtimese Yes, it does if the sled of moving slow enough at the apex. If it is moving fast enough the acceleration would be equal to $g$ and $N=0$. Any faster and the sled will fly off of the hill. – Aaron Stevens Oct 15 at 7:47