Physics, asked by sushant2685, 16 days ago

A car initially at rest  starts moving with a constant acceleration  of 0.5 m / s ^ 2 and travels a distance of 49m . Find its final velocity and the time taken .​

Answers

Answered by Anonymous
4

Required Answer:

• The final velocity of the car = 5 m/s • The time taken by the car = 10 seconds

Given :

• Initial velocity of the car = 0 m/s

• Acceleration of the car = 0.5 m/s² • Total distance travelled by the car = 25 m

To find :

• Final velocity of the car • Time taken by the car

Solution :

To calculate the final velocity of the car, we will use the third equation of motion.

Third equation of motion :

v² - u² = 2as

where,

• v denotes the final velocity

• u denotes the initial velocity

• a denotes the acceleration • s denotes the distance travelled/displacement

Substituting the given values:

→ v² - (0)² = 2(0.5)(25)

→ v² -0 = 2 × 5/10 × 25

→ v² = 5/5 x 25

v² = 25

→ Taking square root on both the sides :

⇒ V = √25

→ V = √(5 × 5)

⇒ V = 5

Therefore, the final velocity of the car = 5 m/s

To calculate the time taken by the car, we will use the first equation of motion.

First equation of motion :

• Vu+ at

where,

• v denotes the final velocity

u denotes the initial velocity

• a denotes the acceleration • t denotes the time taken

Substituting the given values:

5 = 0 + 0.5 x t

→ 50.5 xt

→ 5/0.5 = t

→ 5 × 10/5

⇒ 10 = t

Therefore, the time taken by the car = 10 seconds

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