SATHEE: Kinematics Question 366

Kinematics Question 366

Question: A particle is moving eastwards with a velocity of $5 m s^{- 1}$ . In 10 seconds the velocity changes to $5 m s^{- 1}$ northwards. The average acceleration in this time is

Options:

A) $\frac{1}{2} m s^{- 2}$ toward north

B) $\frac{1}{\sqrt{2}} m s^{- 2}$ toward north-east

C) $\frac{1}{\sqrt{2}} m s^{- 2}$ towards north-west

D) zero

Show Answer

Answer:

Correct Answer: C

Solution:

$Average acceleration = \frac{change in velocity}{time interval}$

$= \frac{Δ \vec{v}}{t}$

$\vec{v_{1}} = 5 \hat{i}, \vec{v_{2}} = 5 \hat{j}$

$Δ \vec{v} = (\vec{v_{2}} - \vec{v_{1}})$

$= \sqrt{v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos 90}$

$= \sqrt{5^{2} + 5^{2} + 0}$

$[A s | v_{1} | = | v_{2} | = 5 m / s] = 5 \sqrt{2} m / s$

$A v g . a c c . = \frac{Δ}{t} = \frac{5 \sqrt{2}}{10} = \frac{1}{\sqrt{2}} m / s^{2}$
$\Rightarrow \tan θ = \frac{5}{- 5} = - 1$ Which means $θ$ it’s in the second quadrant. (Towards north-west)

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Kinematics Question 366

Question: A particle is moving eastwards with a velocity of 5ms−1 . In 10 seconds the velocity changes to 5ms−1 northwards. The average acceleration in this time is

Options:

Answer:

Solution:

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Question: A particle is moving eastwards with a velocity of $5 m s^{- 1}$ . In 10 seconds the velocity changes to $5 m s^{- 1}$ northwards. The average acceleration in this time is