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jmanrules200
06.07.2019 •
Physics
Atoy friction car is propelled forward from rest by a girl's hand which causes it to reach a forward velocity of 3.5m/s in 0.4 seconds. it is then rreleased and later stops in 5.6 seconds. a) what is the acceleration with which the car speeded up? b)what is the acceleration with which the car slowed down? c) how far did the toy car travel?
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Ответ:
a) Acceleration =![8.75m/s^2](/tpl/images/0059/4302/7fd6b.png)
b) Slow down acceleration = 0.625![m/s^2](/tpl/images/0059/4302/adcae.png)
c) Car travels 9.8 m
Explanation:
Acceleration = Change in velocity/Time
a) Change in velocity = 3.5-0 = 3.5 m/s
Time = 0.4 seconds
Acceleration =![3.5/0.4=8.75m/s^2](/tpl/images/0059/4302/51637.png)
b) Change in velocity = 0 - 3.5 = -3.5 m/s
Time = 5.6 seconds
Acceleration =![-3.5/5.6=-0.625m/s^2](/tpl/images/0059/4302/b9421.png)
Slow down acceleration = 0.625![m/s^2](/tpl/images/0059/4302/adcae.png)
c) We have equation of motion,
, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.
Here v = 0 m/s, u = 3.5 m/s, a = -0.625![m/s^2](/tpl/images/0059/4302/adcae.png)
Substituting
So, car travels 9.8 m
Ответ:
Ep = mgh
in your case
37485[J] = 85[kg] * 9.8[m/s^2] * x[m]
the worker is above
x[m]= 45m