Question

wednesday
A CAR
STARS WITH VELOCITY
lom/s
AND ACCELERATES
AT RATE 5 m/s² FIND THE
FINAL VELOCITY
VELOCITY WHEN
THE CAR HAS TRAVELLED
A DISTANCE 30m​

Answer 1

Answer:

20 m/s

Explanation:

Plz check the attachment for your answer.

Plz bro mark my answer as brainliest. It's a request to you brother.

Answer 2

Question :-

A car starts with a velocity of 10 m/s and accelerates at a rate of 5 m/s² . Find the final velocity when the car had travelled a distance of 30 meters .

Answer :-

Given :-

A car starts with a velocity of 10 m/s and accelerates at a rate of 5 m/s²

Required to find :-

• Final velocity of the car after travelling a distance of 30 meters

Equations of motion :-

• 1. v = u + at

• 2. s = ut + ½ at²

• 3. v² - u² = 2as

Solution :-

Given Information :-

A car starts with a velocity of 10 m/s and accelerates at a rate of 5 m/s²

we need to find the final velocity of the car after travelling a distance of 30 meters

From the given information we can conclude that ;

• Initial velocity of the car = 10 m/s

• Acceleration = 5 m/s²

• Displacement = 30 meters

Using the 2nd equation of motion ;

>> s = ut + ½ at²

So,

➦ 30 = 10 x t + ½ x 5 x t²

➦ 30 = 10t + 5½t²

➦ 30 x 2 /5 = 10t

➦ 60/5 = 10t + t²

➦ 12 = 10t + t²

➦ 10t + t² - 12 = 0

By rearranging the terms and factorise the quadratic polynomial

➤ t² + 10t - 12 = 0

➤ t² + 6t - 2t - 12 = 0

➤ t ( t + 6 ) - 2 ( t + 6 ) = 0

➤ ( t - 2 ) ( t + 6 ) = 0

This implies,

=> t - 2 = 0

=> t = 2

Similarly,

=> t + 6 = 0

=> t = - 6

Since,

Time can't be in negative . So, the time taken = 2 seconds

Now,

Using the 1st equation of motion ;

That is ,

• v = u + at

➪ v = 10 + ( 5 )( 2 )

➪ v = 10 + 10

➪ v = 20 m/s

Therefore,

➪ Final velocity ( v ) = 20 m/s