A particle, moving with uniform speed v, changes direction by an angle θ in time t. Magnitude of its average acceleration during this time is

Question

A particle, moving with uniform speed v, changes direction by an angle θ in time t. Magnitude of its average acceleration during this time is (A) zero (B) (2v/t) sin (θ/2) (C) (v√2/t) (D) V/2

Tardigrade Admin · Accepted Answer

aav = | (Δ ν/Δ t) | = (|Vt - Vt|/t) = (√ν2 + ν2 - 2 ν.ν . cos θ/t)

Q. A particle, moving with uniform speed v, changes direction by an angle $θ$ in time t. Magnitude of its average acceleration during this time is

zero

$\frac{2 v}{t} sin \frac{θ}{2}$

$\frac{v 2}{t}$

V/2

$a_{a v} = ∣ ∣ \frac{Δ ν}{Δ t} ∣ ∣ = \frac{∣ V _{t} - V _{t} ∣}{t} = \frac{ν ^{2} + ν ^{2} - 2 ν . ν . c o s θ}{t}$

Q. A particle, moving with uniform speed v, changes direction by an angle θ in time t. Magnitude of its average acceleration during this time is

zero

t2v​sin2θ​

tv2​​

V/2

aav​=∣∣​ΔtΔν​∣∣​=t∣Vt​−Vt​∣​=tν2+ν2−2ν.ν.cosθ​​

Q. A particle, moving with uniform speed v, changes direction by an angle $θ$ in time t. Magnitude of its average acceleration during this time is

$\frac{2 v}{t} sin \frac{θ}{2}$

$\frac{v 2}{t}$

$a_{a v} = ∣ ∣ \frac{Δ ν}{Δ t} ∣ ∣ = \frac{∣ V _{t} - V _{t} ∣}{t} = \frac{ν ^{2} + ν ^{2} - 2 ν . ν . c o s θ}{t}$