A ball is moving in a circular path of radius 5 m. If tangential acceleration at any instant is 10 m/s2 and the net acceleration makes an angle 30 with the centripetal acceleration, then the instantaneous speed is

Question

A ball is moving in a circular path of radius 5 m. If tangential acceleration at any instant is 10 m/s2 and the net acceleration makes an angle 30 with the centripetal acceleration, then the instantaneous speed is (A)  50√3m/s  (B)  9.3m/s  (C)  6.6m/s  (D)  5.4m/s

Tardigrade Admin · Accepted Answer

tan θ =(at/ac)=(at/v2/r) ⇒ v2=(atr/ tan θ ) =(10× 5/ tan 30o)=(10× 5/1/√3) v2=50× √3 =50× 1.732=86.6 v=√86.6=9.3 m/s

Q. A ball is moving in a circular path of radius 5 m. If tangential acceleration at any instant is 10 m/s2 and the net acceleration makes an angle 30 with the centripetal acceleration, then the instantaneous speed is

$503 m / s$

$9.3 m / s$

$6.6 m / s$

$5.4 m / s$

$tan θ = \frac{a _{t}}{a _{c}} = \frac{a _{t}}{v ^{2} / r}$ $\Rightarrow$ $v^{2} = \frac{a _{t} r}{t a n θ}$ $= \frac{10 \times 5}{t a n 30 ^{o}} = \frac{10 \times 5}{1/ 3}$ $v^{2} = 50 \times 3$ $= 50 \times 1.732 = 86.6$ $v = 86.6 = 9.3 m / s$

Q. A ball is moving in a circular path of radius 5 m. If tangential acceleration at any instant is 10 m/s2 and the net acceleration makes an angle 30 with the centripetal acceleration, then the instantaneous speed is

503​m/s

9.3m/s

6.6m/s

5.4m/s

tanθ=ac​at​​=v2/rat​​ ⇒ v2=tanθat​r​ =tan30o10×5​=1/3​10×5​ v2=50×3​ =50×1.732=86.6 v=86.6​=9.3m/s

$503 m / s$

$9.3 m / s$

$6.6 m / s$

$5.4 m / s$

$tan θ = \frac{a _{t}}{a _{c}} = \frac{a _{t}}{v ^{2} / r}$ $\Rightarrow$ $v^{2} = \frac{a _{t} r}{t a n θ}$ $= \frac{10 \times 5}{t a n 30 ^{o}} = \frac{10 \times 5}{1/ 3}$ $v^{2} = 50 \times 3$ $= 50 \times 1.732 = 86.6$ $v = 86.6 = 9.3 m / s$