Position of a particle moving along X-axis is given by x=28t-4t^2. The distance travelled by particle t=0 to t=2 is???
Ans: 8
Anonymous:
Answer is wrong ....
it given in my allen module
answer is 40
plzz check it may be wrong
the options are ...0,8,12,16
and ans is 8
Answers
Answered by
9
Differentiate x (displacement) to get v (velocity)
Then again differentiate v to get a (acceleration)
Then finally use the 2nd equation of motion that is s=ut + 1/2at^2 to find the distance travelled.
Initial velocity at t= 0 is 28m/s
So using 2nd equation of motion
s = 28(2) -1/2×8×4
=56-16
=40
So ans is 40 not 8
HOPE IT HELPS..... :-)
Then again differentiate v to get a (acceleration)
Then finally use the 2nd equation of motion that is s=ut + 1/2at^2 to find the distance travelled.
Initial velocity at t= 0 is 28m/s
So using 2nd equation of motion
s = 28(2) -1/2×8×4
=56-16
=40
So ans is 40 not 8
HOPE IT HELPS..... :-)
Attachments:
i am not sure.....but a is taken into consideration....
instead of initial and final velocity i took acceleration
ya but initial velocity is 28m/s becoz initial velocity is the velocity at t = 0 but u have taken velicity at t=2sec which is instantaneous velocity
ok i got that....
thank you
^_^
so ans is 40
yaa
thank you :-)
wlcm
Answered by
4
Answer:8
Explanation:question is misprinted in Allen's module
It is x=28t-4t^2
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