Answer:
(a) Shortest distance traveled by train = 178.89 m
(b) Constant acceleration will be [tex]a_t=1.118m/sec^2[/tex]
Explanation:
We have given speed v = 72 km /hr = [tex]=\frac{72\times 1000m}{3600sec}=20m/sec[/tex]
Radius of path = 400 m
Total acceleration [tex]a=1.5m/sec^2[/tex]
We know that normal acceleration is given by [tex]a_n=\frac{v^2}{r}=\frac{20^2}{400}=1m/sec^2[/tex]
(b) Total acceleration is given by [tex]a^2=a_n^2+a_t^2[/tex]
[tex]1.5^2=1^2+a_t^2[/tex]
[tex]a_t=1.118m/sec^2[/tex]
So corresponding acceleration will be [tex]a_t=1.118m/sec^2[/tex]
(a) As the train starts from rest its initial velocity u = 0 m/sec
Final velocity v = 20 m/sec
Acceleration [tex]a_t=1.118m/sec^2[/tex]
From third equation of motion we know that [tex]v^2=u^2+2as[/tex]
[tex]20^2=0^2+2\times 1.118\times s[/tex]
s = 178.89 m
The shortest distance in which the train can reach a speed of 72 km/h is equal to 178.89 meters.
Given the following data:
First of all, we would determine the centripetal acceleration of the train by using this formula:
[tex]A_v =\frac{V^2}{r} \\\\A_v =\frac{20^2}{400}\\\\A_v =\frac{400}{400}\\\\A_v =1\;m/s^2[/tex]
The maximum total acceleration experienced by the train is given by:
[tex]A_T^2 =A_v^2 +A_t^2\\\\1.5^2=1^2+A_t^2\\\\A_t=2.25-1\\\\A_t=\sqrt{1.25}\\\\A_t=1.118\;m/s^2[/tex]
For the shortest distance, we would apply the third equation of motion:
[tex]V^2 = U^2 +2aS\\\\20^2 = 0^2 +2(1.118)S\\\\400=2.236S\\\\S=\frac{400}{2.236}[/tex]
S = 178.89 meters.
Read more on distance here: brainly.com/question/10545161
