Distance= speed x time
distance = 2.6 m/s x 120s
distance= 312 meters
Both believe that an atom contains negative charges and positive charges.
But both were different in the placement of charges
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
4200000 is your answer hope this helps
