Question

A particle is moving eastward with a speed of 6m/s,the particle is found to be moving with same speed in a direction 60°north of east.The magnitude of average acceleration in this interval of time is
(1) 6m/s^2 (2)3m/s^2 (3)1m/s^2
(4) Zero

Answer 1

Answer:

The magnitude of average acceleration is 1 m/s².

(3) is correct option.

Explanation:

Given that,

Initial speed $v_{i}= 6\hat{i}\ m/s$

Angle = 60° north of east

Suppose the time is 6 sec.

We need to calculate the final speed

Using component

$v_{f}=6\cos\theta\hat{i}+6\sin\theta\hat{j}$

$v_{f}=6\cos60\hat{i}+6\sin60\hat{j}$

$v_{f}=3\hat{i}+3\sqrt{3}\hat{j}$

We need to calculate the change in velocity

$\vec{\Delta v}=v_{f}-v_{i}$

$\vec{\Delta v}=3\hat{i}+3\sqrt{3}\hat{j}-6\hat{i}$

$\vec{\Delta v}=3\sqrt{3}\hat{j}-3\hat{i}$

The magnitude of the change in velocity

$\Delta v=\sqrt{(3)^2+(3\sqrt{3})^2}$

$\Delta v=6\ m/s$

We need to calculate the acceleration

Using formula of acceleration

$a = \dfrac{\Delta v}{t}$

Put the value into the formula

$a=\dfrac{6}{6}$

$a=1\ m/s^2$

Hence, The magnitude of average acceleration is 1 m/s².

Answer 2

Answer:

answer is given in the attachment