Answer to Question #173188 in Mechanics | Relativity for Maezhei

Question

A trains from rest at station A and it took 150 seconds to reach station B. The speed-time curve of the motion is given below. Find the a) distance of the railway track connecting the stations and b) the acceleration of the train within the time interval 20-80 seconds.

Answer 1

according to given question and some own knowledge (to complete question)

GIVEN:

Initial speed at A= $u=0$

speed at point B $= v_1$

speed at point C $= v_2$

speed at end point D $= v_3=0$

TIME:

from A TO B ($t_1 =20 sec$ )

from B TO C ($t_2 ={840\over13}=64.61sec$ )

from C TO D ($t_3 =70 sec$ )

GRAPH FOR GIVEN QUESTION $\implies$

LET;

ACCELERATION be $a_1, a_2,a_3$ for respective time intervals.

NOW;

we find value of each acceleration,

using first eqn. of motion

$\boxed{v_1=u+a_1t_1}$

$30=0+a_1(20) \\ \boxed{a_1=1.5mi/min^2}$

similarly;

$\boxed{v_2=v_1+a_2t_2}$

$\bigstar$ $64.61=30+a_2(60)\\ \boxed{a_2=0.576mi/min^2}$

similarly;

$\boxed{v_3=v_2+a_3t_3}$

$0=64.61+a_3(70) \\ \boxed{a_3=-0.923mi/min^2}$

b)

NOW;

value of DISTANCE covered in respective time span .

$S=ut+{1\over2}at^2 \\S_1=0+{1\over2}a_1({20\over60})^2 \implies S_1=0.0833 \\ \\S_2={30\over60}* ({60\over60})+{1\over2}a_2({60\over60})^2 \implies S_2=0.7884 \\ \\S_3={64.61\over60}* ({70\over60})+{1\over2}a_3({70\over60})^2 \implies S_3=0.6281$

$\bigstar$ Therefore total distance travelled (displacement) $S_1+S_2+S_3$ =1.5 miles