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4 years ago

7

A pitcher tosses a baseball in the positive direction with a speed of 12 m/s. If the ball stops after 4 seconds, what is the ave

rage acceleration of the ball?
A.) -3 m/s/s
B.) +48 m/s/s
C.) +3 m/s/s
D.) -48 m/s/s

1 answer:

meriva4 years ago

8 0

Answer:

option c is correct

Explanation:

we know that time rate of change of velocity is called acceleration

so a=v/t

a=12/4

a=3m/s²

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3 years ago

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3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

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-24.28571 rad/s²

29.57239 revolutions

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52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

3 years ago