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Sati [7]
3 years ago
10

What is the equation for finding the acceleration of an object moving in a straight line?

Physics
1 answer:
lapo4ka [179]3 years ago
5 0

The equations to find the acceleration are the suvat equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t

Explanation:

The acceleration of an object is the rate if change in velocity of the object. It is calculated as

a=\frac{v-u}{t}

where

v is the final velocity of the object

u is the initial velocity

t is the time elapsed

For an object moving in a straight line at constant acceleration, there are several equations that can be used to find the acceleration: they are called suvat equations. They are the following:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t

where

u is the initial velocity

v is the final velocity

t is the time

s is the distance covered

a is the acceleration

Therefore, any of the above equations can be used to  calculate the acceleration.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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At the intersection of Texas Avenue and University Drive,
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Answer:

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Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

The final linear momentum will be:

\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

so:

 \vec{P}_0= \vec{P}_f 

( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

so

\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

So, for the truck

m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

And, for the car

950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

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