Question

rest at the rate of 4 ms -2. a ) what will be the velocity attained in 10 second b) how far will it travel in this duration?

Answer 1

Given :-

• Initial velocity, u = 0 m/s
• Acceleration, a = 4 m/s²

To determine :-

a) The velocity attained in 10 seconds.

b) The distance travelled in this duration.

Solution :-

a) • u = 0 m/s

• a = 4 m/s²

• t = 10 secs

We know,

$\bigstar\ \color{green} {a=\dfrac{v-u}{t} }$

Putting the values :-

$\sl{\hookrightarrow 4=\dfrac{v-0}{10} }\\\\\sl{\hookrightarrow 4=\dfrac{v}{10} }\\\\\boxed{\color{maroon}{\hookrightarrow v=40\ m/s }}$

Hence, the velocity after 10 s will be 40 m/s.

$\rule{190}3$

b) • u = 0 m/s

• a = 4 m/s²

• v = 40 m/s

We know,

$\bigstar\ \color {green} {v^2=u^2+2as }$

Putting the values :-

$\sl{\hookrightarrow (40)^2=(0)^2+2\times4\times s}\\\\\sl{\hookrightarrow 1600=0+8s}\\\\\sl{\hookrightarrow 8s=1600}\\\\\sl{\hookrightarrow s=\dfrac{1600}{8} }\\\\\boxed{\color{maroon}{\hookrightarrow s=200\ m}}$

Answer 2

Correct Question -

A body starts from rest with an accelaleration of 4 m / s².

To Find -

[ A ] What will be the velocity attained in 10 second ?

[ B ] How far will it travel in this duration ?

Solution -

In the above question , the following information is given :

A body starts from rest with an accelaleration of 4 m / s².

As the body starts from rest , the innitial Velocity of the body is 0

Now ,

The acceleration of the body is 4 m / s² .

So ,

According to the first law of motion ,

V = u + at

$\sf{ V_{10} = 0 + 4 \times 10 } \\ \\ \sf{ \implies { V_{10} = 40 m / s^2 }$

Now ,

According to the second Equation of motion -

S = ut + ( 1 / 2 ) a t²

=> S = ( 1 / 2 ) × 4 × 100

=> 200 m

Answer -

[ A ] 40 m / s¹

[ B ] 200 m .

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