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3 years ago

6

What is the term for the overall change in position?

2 answers:

valentina_108 [34]3 years ago

8 0

Hi!

When we're looking at a basic overview of a change in position, this is known as <em>displacement. </em>Displacement is a vector quantity, compared to distance, which is a scalar quantity.

What this means is that displacement is focusing on magnitude <em>and </em>direction<em />, instead of just the magnitude.

Hopefully, this helps! =)

mart [117]3 years ago

6 0

That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.

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3 years ago

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2 years ago

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2 years ago

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2 years ago

Read 2 more answers

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago