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atroni [7]
3 years ago
10

A car sets off from the traffic lights. It reaches a speed of 27 m/s in 18 seconds. What is its acceleration?

Physics
1 answer:
Katarina [22]3 years ago
6 0

Answer:

1.6875m/s²

Explanation:

acceleration =  \frac{final \: velocity \:  -  \: initial \: velocity}{time \: taken}

<em> </em><em>a </em><em>=</em><em> </em><em>unknown.</em>

<em>v </em><em>(</em><em>final </em><em>velocity)</em><em> </em><em>=</em><em> </em><em>2</em><em>7</em><em>m</em><em>/</em><em>s</em>

<em>u </em><em>(</em><em>initial </em><em>velocity</em><em>)</em><em> </em><em>=</em><em> </em><em>0</em><em>m</em><em>/</em><em>s </em><em>(</em><em>as </em><em>the </em><em>car</em><em> </em><em>was </em><em>stationary </em><em>in </em><em>the </em><em>traffic </em><em>lights)</em><em>.</em>

<em>t </em><em>(</em><em>time </em><em>taken)</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>s</em>

<em>a =  \frac{27 - 0}{16}</em>

<em>acceleration</em><em> </em><em>=</em><em> </em><em>1.6875</em><em>m</em><em>/</em><em>s²</em>

hope it helps. :)

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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
3 years ago
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