Question 135: Motion in 2D & Vectors

Question

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity
changes to 5 m/s northwards. The average acceleration in this time is

Accepted Answer

Average acceleration
$$ = {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{m int}} erval}}$$
$$ = {{\Delta \overrightarrow v } \over t}$$
$${\overrightarrow v _1} = +5\widehat i$$, towards east direction.
$$\overrightarrow {{v_2}}  = +5\widehat j$$, towards north direction
$$\Delta \overrightarrow v  = \overrightarrow {{v_2}}  - \overrightarrow {{v_1}} $$ = $$5\widehat i - 5\widehat j$$
$$	herefore$$ $$\,\,\,\,\overrightarrow a  = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$$
$$	herefore$$ $$\,\,\,\, a  = {{\sqrt {{1^2} + {{\left( { - 1} ight)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$$
$$	an 	heta  = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$$
$$	herefore$$ $$\,\,\,\,	heta  = {45^ \circ }$$
Therefore the direction is North-west.

Answer

$${1 \over 2}m{s^{ - 2}}$$ towards north

Answer

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$ towards north-east

Answer

$${1 \over {\sqrt 2 }}m{s^{ - 2}}$$  towards north-west

Answer

zero

Motion in 2D & Vectors

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

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