A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is

Question

A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is (A) Zero (B) (1/√2) m/s2 N-W (C) (1/√2) m/s2 N-E (D) (1/√2) m/s2 S-W

Tardigrade Admin · Accepted Answer

Δ v = 2v sin ((θ/2)P) = 2× 5 × sin 45° = (10/√2) ∴ a = (Δ v/Δ t) = (10/√2/10) = (1/√2) m/s2

Q. A particle moves towards east with velocity $5 m / s$ . After $10$ seconds its direction changes towards north with same velocity. The average acceleration of the particle is

Zero

$\frac{1}{2} m / s^{2} N - W$

$\frac{1}{2} m / s^{2} N - E$

$\frac{1}{2} m / s^{2} S - W$

$Δ v = 2 v sin (\frac{θ}{2} P) = 2 \times 5 \times sin 4 5^{\circ}$

$= \frac{10}{2}$

$∴ a = \frac{Δ v}{Δ t}$

$= \frac{10/ 2}{10} = \frac{1}{2} m / s^{2}$

Q. A particle moves towards east with velocity 5m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is

Zero

2​1​m/s2N−W

2​1​m/s2N−E

2​1​m/s2S−W

Δv=2vsin(2θ​P)=2×5×sin45∘ =2​10​ ∴a=ΔtΔv​ =1010/2​​=2​1​m/s2

Q. A particle moves towards east with velocity $5 m / s$ . After $10$ seconds its direction changes towards north with same velocity. The average acceleration of the particle is

$\frac{1}{2} m / s^{2} N - W$

$\frac{1}{2} m / s^{2} N - E$

$\frac{1}{2} m / s^{2} S - W$

$Δ v = 2 v sin (\frac{θ}{2} P) = 2 \times 5 \times sin 4 5^{\circ}$

$= \frac{10}{2}$

$∴ a = \frac{Δ v}{Δ t}$

$= \frac{10/ 2}{10} = \frac{1}{2} m / s^{2}$