Explanation:
It is given that,
Mass of the car 1,
Initial speed of car 1, (east)
Mass of the car 2,
Initial speed of car 2, (north)
(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :
The magnitude of speed,
V = 12.22 m/s
(b) Let is the direction the wreckage move just after the collision. It is given by :
Hence, this is the required solution.
Answer:
Explanation:
This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is
which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):
and get
12.1 m
Answer: a) 95.07m b) 81.88 m
Explanation:
a)
For finding the distance when vehicle is going downhill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 +
Stop sight distance= 95.07 m
b)
For finding the distance when vehicle is going uphill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 +
Stop sight distance= 81.88 m
