The parallel component is given by
F=180cos(25)=163.14N
Hi there!
Acceleration = change in velocity / change in time = Δv/Δt
Thus:
a = (75 - 15)/4 = 60/4 = 15 mi/hr²
Answer:
A.) 1430 metres
B.) 80 seconds
Explanation:
Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:
U = acceleration × time
U = 1.1 × 20 = 22 m/s
It then proceeds at constant speed for 1100 m
Then, time t will be
Time = distance/ velocity
Time = 1100/22
Time = 50 s
before slowing down at 2.2m/s^2 until it stops at the station.
Deceleration = velocity/time
2.2 = 22/t
t = 22/2.2
t = 10s
Using area under the graph, the distance between the two stations will be :
(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)
220 + 1100 + 110
1430 m
The time taken between the two stations will be
20 + 50 + 10 = 80 seconds
