from objective 2 que no.10

The accelerations of a particle as seen from two frames S₁ and S₂ have equal magnitude 4m/s².

(a)The frames must be at rest with respect to each other.

(b)The frames may be moving with respect to each other but neither should be accelerated with respect to each other.

(c)The acceleration of S₂ with respect to S₁ may either be zero or 8m/s².

(d)The acceleration of S₂ with respect S₁ may be anything b/w zero and 8m/s².

ans-d

i need explanation

Hi Suchita,

As acn of particle (say p) wrt frames S1 and S2 has mag of 4m/s^2.

We write A_p/s1 = 4*[x1(i) + y1(j)]. Where x1(i)+y1(j) is a unit vector.

And similarly A_p/s2 = 4*[x2(i) + y2(j)]. Where x2(i) + y2(j) is another unit vector.

So now A_s2/s1 = A_p/s1 - A_p/s2 ----------- [this is because acn A_p/s1 = A_p - A_s1, in the vector notation]

so ie = 4[unit vec 1 - unit vec 2] = 4[a-b] {where unit vectors are denoted by a and b}.

So the mag of acn is |A_s2/s1| = 4|a-b| ----------(Note that |a|

Consider |a-b|² = (a-b)dot(a-b) = |a|²+|b|²-2(|a|*|b|*cosx) = 2-2cosx = 2(1-cosx) = 2*2*sin²x/2 = 4sin²x/2

So |a-b| = |2sinx/2|.

And hence |A_s2/s1| = 8|sin(x/2)|

And clearly |sinx/2| lies between 0 to 1. And hence answer (D).

Also note, the above method is a general approach. For special cases of S1, and S2, and the particle moving along the same straight line, angle between a and b could be 0 or 180. And hence x/2 can be 0 or 90. (In this case the answer would be (C)).

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).

Approved

consider a stationary particle.

s1 is moving with yhat particle with a=4ms-²

s2 is moving with a=4ms-²

both frames r moving with an angleΦ now Φ can lie btween 0 to 180 so they can move wrt each other with minimum 0 relative acc.maximum or 8ms-^{2 .}