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2 years ago

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A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this

time
find distance and acceleration

PLEASEEEE HELP I HAVE A PHYSICS EXAM TOMORROW AND IM STRUGGLING

1 answer:

krok68 [10]2 years ago

8 0

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

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Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

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