The path of a particle moving in the x-y plane [v (2 s) > v (1 s) > v (0 s)] is

Question

The path of a particle moving in the x-y plane [v (2 s) > v (1 s) > v (0 s)] is shown by dotted line in the figure. At t = 1.0 s, the particle is at A. If the particle is slowing down when it reached point A [that means v (1 s) < v (0 s)], what is the direction of acceleration (instantaneous) vector? The normal to the curve at A is indicated by red line. Choices:

To the right, along the normal

To the left, along the normal

a.

To the right, between the normal (red line) and at 1 s.

b.

To the left, between the normal (red line) and   at 1 s.

c.

To the right, along the normal

d.

To the right, between the normal (red line) and   at 1 s.

e.

To the left, between the normal (red line) and at 1 s.

f.

To the left, along the normal

y (m) D, Acceleration in two dimensional motion 2.0 t 1.0 1.0 Rover's path (m) Ol 0.5 I 0 1.5 2.0

Explanation / Answer

>> In a Curvature, there are two types of acceleration, experienced by the particle . These are :

1). Centrepetal Aceleration, Ac : It is aong the normal to the path and directd towards center.

So, here, Ac will be directed towards left side and normal to the path ....

2). Tangential Acceleration, At : It acts tangentially to the path and is responsible for change in speed.

Now, here, as speed is decreasing, So, At will act opposite to the velocity direction and so,

At acts tangentially to the path downward side...

>> Now, as Total Acceleration is the resultant of these two accelerations : Ac and At

So, there resultant will lie on the left side between the normal (red line) and minus velocity ....

Hence, OPTION (B) is correct ....

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