yashveer singh
class 11th
v(0)=10m/s
uniform acc..=-2m/s2
dv / dt=-2
dv=-2dt
∫dv=-2∫dt
v=-2t+c ----eqn(1)
At t=0,v=10m/s----
10=-2*0+c
c=10
putting the value of c in eqn(1).....
v=-2t+10 --------It is the v-t eqn
now let’s find the time when v=0,
from v-t eqn-----
0=-2t+10
t=5sec (At t=5sec v=0)
now finding the s-t eqn from the v-t eqn-----
by integration.....
v=-2t+10
ds / dt=-2t+10
ds=-2tdt+10dt
∫ds=-2∫tdt+10∫dt
s=-t2+10t --------Its the s-t eqn
now finding displacement in t=5sec when v=0,
s=-(5)2+10*5 (from s-t eqn)
s=25m
let say this displacement as D1
D1=25m
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v(5)=0
uniform acc..=-2m/s2
finding v-t eqn.........
dv / dt=-2
dv=-2dt
by integation
v=-2t+c
At t=5,v=0;
v=-2t+10
ds / dt=-2t+10
ds=-2tdt+10dt
now by integrating...
s=-t2+10t
at t=12’displkacement=?,
s=-(12)2+10*12
s=-144+120
s=-24
lets say this displacement as D2
D2=-24m
now total distance=|D1|+|D2 |
25+24=49m
now total displacement=D1+D2
25+(-24)=1m
HOPE THIS ANSWER CLEARS TJHE CONCEPT OF SOLVING THE QUESTION BY DIFFERENTIATION AND INTEGRATION.........................................
