QUESTION 1 The path of a particle moving in the x-y plane [v (2 s) > v (1 s) > v
ID: 1415959 • Letter: Q
Question
QUESTION 1
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.
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, between the normal (red line) and at 1 s.
d.To the right, along the normal
e.To the left, along the normal
f.To the left, between the normal (red line) and at 1 s
Explanation / Answer
As, instantaneous acceleration = d2r/dt2
= dv/dt
=> the direction of acceleration (instantaneous) vector = To the left, between the normal (red line) and minus V with rightwards arrow on top at 1 s.
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