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3 years ago

If a car travels 600m west in 25 seconds, what is its velocity?

Physics

1 answer:

Elena L [17]3 years ago

4 0

$v = \frac{ \:the covered distance}{time} \\$

$v = \frac{600}{25} = \frac{6 \times 100}{25} = 6 \times 4 = 24 \: \\$

$v = 24 \: \frac{m}{s} \\$

_________________________________

If west means the west of the axis x the velocity equal :

$- 24 \: \frac{m}{s} \\$

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Read 2 more answers

A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

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Answer:

$V_f = 13.8863 \angle 60.89\°$

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We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

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For the particular case on the Y axis, we do the same with the speed of object 1.

$m_2v_2=(m_1+m_2)v_fsin\theta$

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

$Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°$

Replacing in any of the two functions, given above, we will find the final speed after the collision,

$(950)(16)=(950+1300)V_fcos(60.89)$

$V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}$

$V_f = 13.8863 \angle 60.89\°$

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