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# A particle moving with constant speed turns left by an angle of 74° after travelling every 1 m distance. If it returns back to its staring pt. after 18 sec , then the magnitude of average acceleration of the particle frm its initial posn to just bfore it takes its 2nd turn is 10 k m/s2 . Find k.

Ashutosh
16 Points
one month ago
Let number of turns be m, so 74°×m=360°×n i.e a multiple of 360°.
So, m = 180
Therefore speed = 180×1/18=10m/s
Now average acceleration,
a = (v1-v2)/(t1-t2) = V/T
V = 10 - 10(cos 74° i + sin 74° j)
So, magnitude of V = 12
Now time taken for two turns = (18/180) ×2 = 1/5 seconds
So, a = V/T = 12/(1/5) = 60 m/s²
So, k = 6