Question

A car with an initial speed of 4.30 m/s accelerates

uniformly at the rate of 3.00 m/s2. What is the car's speed
after 5.00 s? How far did it travel in that time?

Answer 1

Heya!!

For calculate final velocity, lets applicate formula

`\boxed{V=V_o+a*t}`

Δ Being Δ

V = Final Velocity = ?

Vo = Initial velocity = 4,3 m/s

a = Aceleration = 3 m/s²

t = Time =5 s

⇒ Let's replace according the formula:

`\boxed{V=4,3\ m/s +3\ m/s* 5\ s}`

⇒ Resolving

`\boxed{V=19,3\ m/s}`

Result:

The velocity after 5 sec is 19,3 meters per second (m/s)

For calculate distance, lets applicate formula

`\boxed{x=V_o*t+(a*t^(2))/(2)}}`

Δ Being Δ

x = Distance = ?

Vo = Initial velocity = 4,3 m/s

a = Aceleration = 3 m/s²

t = Time =5 s

⇒ Let's replace according the formula:

`\boxed{x=4,3 * 5 +(3*5^(2))/(2)}}`

⇒ Resolving

`\boxed{x = 59 \ m}`

Result:

The distance is 59 meters.

Good Luck!!