Spreads out in the medium, first choice
(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s
Acceleration a = 0.0600 m/s²
time = 8 min = 8 x 60 = 480 s
Final velocity
= u + at
= 5.00 + 0.0600(480)
= 33.8 m/s
The final velocity is 33.8 m/s
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
You didn't include the numerical value of speed.
