Answer:
No, there won't be a collision.
Explanation:
We will use the constant acceleration formulas to calculate,
v = u + a*t
0 = 25 + (-0.1)*t
t = 250 seconds (the time taken for the passenger train to stop)
v^2 = u^2 + 2*a*s
0 = (25)^2 + 2*(-0.1)*s
s = 3125 m (distance traveled by passenger train to stop)
If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur
Speed*time = distance
Distance = (15)*(250)
Distance = 3750 m
As the distance is way more, there won’t be a collision
Answer:
10 m/s^2
Explanation:
Equation: F = ma.
a = acceleration
m = mass
F = force
Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.
F = 20
m = 2
a = ?
a = F/m
a = 20/2
a = 10 m/s^2
Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
86.3 I just did the math and that’s the answer I got
