Speed with which initially car is moving is 21 m/s
Reaction time = 0.50 s
distance traveled in the reaction time d = v t
d = 21 * 0.50 = 10.5 m
deceleration after this time = -10 m/s^2
now the distance traveled by the car after applying bakes
so total distance moved before it stop
d = 22.05 + 10.5 = 32.55 m
so the distance from deer is 35 - 32.55 = 2.45 m
now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop
so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m
again by kinematics
so maximum speed would be 22.1 m/s
Answer:
balanced , balanced , unbalanced, unbalanced, balanced, balanced, unbalanced
Explanation: you're welcome
The trains take <u>57.4 s</u> to pass each other.
Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.
therefore,
The relative velocity of the train A with respect to B is given by,
If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.
The trains are a distance d = 2.71 km apart.
Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,
Substitute 2.71 km for d and 170 km/h for 
Express the time in seconds.
Thus, the trains cross each other in <u>57.4 s</u>.
400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent
