Question

A body moving with uniform acceleration covers a displacement 12 metre in the first 4 second interval if it starts from rest find the displacement covered by the same body in the next 6 second interval​

Answer 1

Answer:

distance travelled by in next 6 seconds

= 63 meters

Step by step explanations :

given that

A body moving with uniform acceleration covers

a displacement 12 metre

in the first 4 second

and

given,

it starts from rest

so,

initial velocity of the body = 0 m/s

displacement of the body = 12 meters

time taken to accelerates = 4 seconds,

let the final velocity of the body be v

now we have,

initial velocity(u) = 0 m/s

final velocity = v

displacement(s) = 12 m

time taken(t) = 4 seconds

by the equation of motion,

S = ut + ½ at²

where,

a is the acceleration,

putting the values,

12 = 0(4) + ½ a × (4)(4)

12 = 8a

a = 12/8

a = 3/2

a = 1.5 m/s²

also,

by the equation of motion,

v = u + at

again putting the values,

v = 0 + (1.5)4

v = 6 m/s

so,

velocity of the body after 4 seconds

= 6m/s

now,

to find,

displacement covered by the same body in the next 6 second interval

so,

here,

initial velocity = velocity after 4 seconds

= 6 m/s

so,

initial velocity(u) = 6 m/s

acceleration(a) = 1.5 m/s²

time to be taken(t) = 6 seconds

S = ut + ½ at²

S = (6)(6) + ½ (1.5)(6)(6)

S = 36 + 27

S = 63

so,

distance travelled by in next 6 seconds

= 63 meters

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\huge {\boxed {\bold {\color {red} {63 m}}}}$

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Given—

starts from rest

initial velocity = 0 m/s

displacement = 12 meters

time taken to accelerates = 4 seconds,

let the final velocity be v

To find—

Distance travelled in next 6 sec

By NLM

S = ut + ½ at²

Substituting the value

12 = 0(4) + ½ a × (4)(4)

12 = 8a

a = 12/8

a = 3/2

a =1.5m/s²

by Newton 1st law

v = u + at

v = 0 + (1.5)4

v = 6 m/s

so,

velocity of the body after 4seconds

=6m/s

To find displacement covered by the same body in the next 6 second

so,

here,

initial velocity = velocity after 4 seconds

= 6 m/s

so,

initial velocity(u) = 6 m/s

acceleration(a) 1.5 m/s²

time to be taken(t) = 6 seconds

Again by NLM

S = ut + ½ at²

S = (6)(6) + ½ (1.5)(6)(6)

S = 36 + 27

S = 63

Therefore distance travelled by in n 6 sec= 63 meters

$\huge{\green{\ddot{\smile}}}$