Answer:
22.2 m/s
Explanation:
First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.
Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.
The average speed can be found by using the equation . After substitution, this gives the fraction
, which reduces to 22
m/s, or about 22.2 m/s.
Recall this kinematic equation:
a =
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a =
a) The train collide after 22.5 seconds
b) The trains collide at the location x = 537.5 m
c) See graph in attachment
d) The freight train must have a head start of 500 m
e) The deceleration must be smaller (towards negative value) than
f) The two trains avoid collision if the acceleration of the freight train is at least
Explanation:
a)
We can describe the position of the passenger train at time t with the equation
where
is the initial velocity of the passenger train
is the deceleration of the train
On the other hand, the position of the freight train is given by
where
is the initial position of the freight train
is the constant velocity of the train
The collision occurs if the two trains meet, so
This is a second-order equation that has two solutions:
t = 22.5 s
t = 177.5 s
We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.
b)
In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.
The position of the freight train is
And substituting t = 22.5 s, we find:
We can verify that the passenger train is at the same position at the time of the collision:
So, the two trains collide at x = 537.5 m.
c)
In the graph in attachment, the position-time graph of each train is represented. We have:
The two trains meet at t = 22.5 s, where the position is 537.5 m.
d)
In order to avoid the collision, the freight train must have a initial position of
such that the two trains never meet.
We said that the two trains meet if:
Re-arranging,
Substituting the values for the acceleration and the velocity,
The solution of this equation is given by the formula
The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore
Therefore, the freight train must have a head start of 500 m.
e)
In this case, we want to find the acceleration of the passenger train such that the two trains do not collide.
We solve the problem similarly to part d):
Re-arranging
Substituting,
The solution to this equation is
Again, the two trains never meet if the discriminant is negative, so
So, the deceleration must be smaller (towards negative value) than
f)
In this case, the motion of the freight train is also accelerated, so its position at time t is given by
where is the acceleration of the freight train.
Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,
Re-arranging,
And the solution is
Again, the two trains avoid collision if the discriminant is negative, so
Learn more about accelerated motion:
#LearnwithBrainly
