What will be the anser?

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2 Answers
Aug 11, 2018

#5.7 m/s#

Explanation:

Here,considering speed to be the magnitude of linear velocity.

So,if it is moving with constant speed of #v# at a certain point of time,then its Centripetal acceleration is #|vec a_c|=v^2/r# where #r# is the radius of the circular path.

If its linear acceleration is #vec a_r# then,

#vec a_c + vec a_r =vec a#

Given, #|vec a|=15 ms^-2#

Again angle between #vec a_c# and #vec a_r# is #90^@#

So, #a^2=a_r^2 + a_c^2#

Or, #a^2=a_r^2 + v^4/r^2... 1#

Given, #a=15 ms^-2,r=2.5m#

If, #vec a# makes an angle of #30^@# w.r.t #vec a_c# then,

#tan 30=|vec a_r| /|vec a_c|#

I.e #1/sqrt(3) =a_r/(v^2/r)#

Given, #r=2.5m#

So, putting the values and arranging we get,

#a_r =v^2/(2.5 sqrt(3))#

Putting this value of #a_r# in the previous equation #1# we get,

#v=5.7 m/s#

Aug 11, 2018

Slightly shorter approach, same answer

Acceleration vector in non-uniform circular motion is:

  • #bba(t) = underbrace(r dot theta \ bbhat e_theta)_("tangential ") - underbrace(v^2/r\ bb hat e_r)_("inward radial: " = bba_r) #

In given geometry:

# abs(bba_r) = abs(bba ) cos 30 #

#implies v^2 /2.5 = 15 sqrt3/2#

#:. "speed" = abs(bbv) ~~ 5.7 " m/s"#