A particle experiences constant acceleration for 20 seconds after starting from rest. if it travels a distance S1 in first 10 seconds and distance S2 in next 20 seconds then find relation between S1 and S2.
Answer with best explanation will be marked as brainliest
Answers
Answered by
1
u=0
for the distance travelled in first 10 sec
S1=ut+½a(t)²
S1=0×10+ ½a (10)²
S1=50a
speed of the body after 10 seconds will be
v=u+at
v=0+10a
v=10a
now this speed will act as the initial speed for the next ten seconds
ú=10a
S2=út+½a(t)²
S2=10a+½a(10)²
S2=60a
so we have S1=50a and S2=60a
for the distance travelled in first 10 sec
S1=ut+½a(t)²
S1=0×10+ ½a (10)²
S1=50a
speed of the body after 10 seconds will be
v=u+at
v=0+10a
v=10a
now this speed will act as the initial speed for the next ten seconds
ú=10a
S2=út+½a(t)²
S2=10a+½a(10)²
S2=60a
so we have S1=50a and S2=60a
Similar questions