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 2^{nd} 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 at^{2}

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 at^{2}

⇒ S = 5ms^{-1} × 3s + 1/2 × 12ms^{-2} × 3s × 3s

⇒ S=15m+ 54m

⇒ S = 69 m

From 2^{nd} row, a = ?

⇒ S=ut + 1/2 at^{2}

⇒ 92m = 7ms^{-1} × 4s + 1/2 × a × 4s ×4s

⇒ 92 m = 28 m + a × 8s^{2}

⇒ a × 8s^{2} = 92 – 28 m

⇒ a × 8s^{2} = 64 m

⇒ a= 8 m/s^{2}

Now, for the third table,

From second equation of motion,

v^{2} = u^{2}+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 2^{nd} row ⇒ 10= u^{2}+ 2× 5× 8.4

⇒ 10=u^{2}+84

⇒ -74=u^{2}

⇒ u=-8.6m/s

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