Question

a car initially at rest start with a constant acceleration of 0.5m/s and travel a distance of 25m. calculate its final velocity​

Answer 1

Answer:

5 m / sec

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Acceleration ( a ) = 0.5 $m/sec^2$

Distance ( s ) = 25 m

We have to find final velocity ( v ).

From third equation of motion we have

$\large v^2=u^2+2as$

where  v = final velocity

u = Initial velocity

a = Acceleration

s = Distance

Put the values in equation we get

$\large v^2=0^2+2\times25\times0.5$

$\large v^2=50\times0.5\\\\\\\large v^2=25\\\\\\\large v=\sqrt{25}$

v = 5 m / sec.

Thus the final velocity is 5 m / sec

Answer 2

Answer:

5 m/sec

Explanation:

Given Problem:

A car initially at rest start with a constant acceleration of 0.5m/s and travel a distance of 25m.Calculate its final velocity​.

Solution:

To Calculate:

Its final velocity.

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Method:

Velocity:

The speed of something in the given direction.

Initial velocity:

Initial velocity,vi is the velocity of the object before acceleration causes a change.

Acceleration:

A vehicle's capacity to gain speed.

Distance:

The space between 2 ponits called distance.

Given that,

= > Initial velocity ( u ) = 0 m/sec

= > Acceleration ( a ) = 0.5 m/sec²

= > Distance ( s ) = 25 m

We know that,

Third equation of Motion:

⇒v² = u² + 2as

Here,

⇒v = final velocity.

⇒u = Initial velocity.  

⇒a = Acceleration.

⇒s = Distance.

Now plug values in equation,

⇒ v² = 0² + 2 × 25 × 0.5

⇒ v² = 50 × 0.5

⇒ v² = 25

⇒ v = √25

So,

⇒ v = 5 m / sec.

Hence,

The final velocity is 5 m/sec.