A car moved initially with constant speed u=12m/s on a horizontal road.
Then the driver shut down the engine so that the car decelerated. The
car has a mass of 400KG and the frictional force on the car is 1600N.
Find out the displacement 6s after the engine was shut down.
ANS is 18m
PHYSICS Mechanics
2007-09-21 2:10 am
回答 (3)
2007-09-21 2:19 am
✔ 最佳答案
By Newton 2nd law of motion,f = ma
-1600 = 400a
a = -4 m/s^2 (-ve acceleration means deceleration)
Initial speed, u = 12 m/s
Acceleration, a = -4 m/s^2
Final velocity, v = 0
So, by v^2 = u^2 + 2as
(0)^2 = (12)^2 + 2(-4)s
s = 18 m
By s = (u + v)t/2
18 = (12 + 0)t / 2
t = 3 s
So, after 3 s the engine shut down, the car stops and has a displacement of 18 m.
Therefore, at 6 s after the engine was shut down, the displacement of the car is 18 m.
參考: Myself~~~
2007-09-21 2:22 am
v^2-u^2=2as
12^2-0^2=2(1600/400)(s)
s=18m
1600/400 is becoz :
F=ma
a=F(force)/m(mass)
=1600/400
12^2-0^2=2(1600/400)(s)
s=18m
1600/400 is becoz :
F=ma
a=F(force)/m(mass)
=1600/400
2007-09-21 2:18 am
deceleration= 1600N/400KG =4m/s
time for car to get rest: 12/4 = 3s
displacement= area under the graph speed vs time
12m/s x 3s /2 =18m
time for car to get rest: 12/4 = 3s
displacement= area under the graph speed vs time
12m/s x 3s /2 =18m
收錄日期: 2021-04-13 13:33:48
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