Question

A body covers 4 m in 3rd sec and 12 min 5 sec
in uniformly accelerated motion. The uniform
acceleration of the body is
1) 4 m/s2
2) 2 m/s2
3) 6 m/s2
4) 8 m/s2


WITH EXPLANATION​

Answer 1

GIVEN :

• A BODY COVERS 4m in 3ʳᵈ SEC AND 12m IN 5ᵗʰ SEC

TO FIND :

• UNIFORM ACCELERATION OF BODY

SOLUTION :

FROM EQUATION OF MOTION ,

Sₙ = u + a [ (2n - 1)/2 ]

HERE Sₙ IS DISPLACEMENT IN nᵗʰSEC , u IS INITIAL VELOCITY , a IS ACCELERATION AND n IS THE PARTICULAR SEC

GIVEN THE BODY COVERED 4 m IN 3ʳᵈ SEC

=> 4 = u + a [ ( 2n-1) / 2 ]

=> 4 = u + a [ (2×3-1)/2 ]

=> 4 = u + a [ ( 5/2) ]

=> a = (4 - u) / (5/2)

=> a = 8 - 2u / 5 ----------(1)

ALSO GIVEN THE BODY COVERED 12 m IN 5ᵗʰ SEC

=> 12 = u + a [ (10-1) / 2 ]

=> 12 = u + a [ 9/2 ]

=> a = 12-u / (9/2)

=> a = 24 - 2u / 9 ---------(2)

HERE THE BODY HAS UNIFORM ACCELERATION SO EQ(1) AND EQ(2) CAN BE EQUATED ,

=> 8-2u / 5 = 24-2u /9

=> 9 ( 8 - 2u ) = 5( 24 - 2u )

=> 72 - 18 u = 120 - 10u

=> -8u = 48

=> u = -6 m/s

FROM EQ(1)

a = 8 - 2u / 5 = [ 8 - 2(-6) ] / 5

= [ 20/5 ] = 4 m/s²

∴ ACCELERATION OF BODY IS 4 m/s²

[ NOTE : HERE EQUATIONS OF MOTION ARE VALID SINCE THE BODY HAS UNIFORM ACCELERATION ]

Answer 2

Answer:

\huge\rm\underline\pink{Solution:-}

Solution:−

Mass (m) = 180g = 0.18kg

Volume (v) = 15cm³ = 0.15 m³

\large\rm\bold{\boxed{Density\:=\:\frac{Mass}{Volume}}}

Density=

Volume

Mass

i] Density = 0.18/0.15 = 1.2kg/m³

Relative density of aluminium with respect to which should we calculate ?