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stepan [7]
3 years ago
11

A car is stopped at a red light. When the light turns green, the car accelerates until it reaches a final velocity of 45 m/s. It

takes the car 12 s to reach this speed. How far does the car travel during this time.
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

Speed at the beginning  =  zero

Speed after 12 seconds  =  45 m/s

Average speed during the 12 seconds  =  (1/2) · (0 + 45)  =  22.5 m/s

Distance covered in 12 seconds at an average speed of 22.5 m/s  :

Distance = (speed) x (time)

Distance = (22.5 m/s) x (12 sec)

<em>Distance   =  270 meters </em>  

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
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v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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3 years ago
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