Q. 34.2( 45 Votes )

Complete the following table.


Answer :


From the first equation of motion,


v=u+at


here,


v = final velocity


u = initial velocity


a is the acceleration


t is the time


In the first table,


From Ist row, v = ?


v =2m/s + 4ms-2 × 3s


v = 2m/s + 12 m/s


v = 14 m/s


from 2nd row, u = ?


v = u + at


20 m/s = u + 5ms-2 × 2s


20 m/s – 10m/s = u


u = 10 m/s


Now,


In the second table,



From the equation of motion,


S=ut + 1/2 at2


Where, s is the distance


“u” is the initial velocity.


“v” is the final velocity.


“t” is the time.


“a” is the acceleration


From Ist row, s = ?


S=ut + 1/2 at2


S = 5ms-1 × 3s + 1/2 × 12ms-2 × 3s × 3s


S=15m+ 54m


S = 69 m


From 2nd row, a = ?


S=ut + 1/2 at2


92m = 7ms-1 × 4s + 1/2 × a × 4s ×4s


92 m = 28 m + a × 8s2


a × 8s2 = 92 – 28 m


a × 8s2 = 64 m


a= 8 m/s2


Now, for the third table,



From second equation of motion,


v2 = u2+2as


From Ist row 8 = 4 × 4 + 2 × 3 × S


8 = 16 + 6 × S


-8 = 6 × S


-8/6= S


S= -3/2 =-1.5 m


From 2nd row 10= u2+ 2× 5× 8.4


10=u2+84


-74=u2


u=-8.6m/s


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