# Determining velocity after acceleration over a certain distance?

A car passes you at 18 m/s to the north and increases its speed at a rate of 3.0 m/s^2. Determine the car's velocity when it has accelerated for 50m.

I used the formula Vf^2 = Vi^2+2ad and my answer was more then 15m off the final answer.

What am i doing wrong !?

### 3 Answers

- AnthonyLv 78 years agoFavourite answer
v^2 = u^2 + 2as

v^2 = 18^2 + 2*3*50

v^2 = 324 + 300 = 624m^2/s^2

v = 24.98 m/s = 25 m/s to 2 sig figs.

Assuming the acceleration was in the same direction of travel, the car is still moving north of you.

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- sharlLv 44 years ago
in the airplane reference physique, preliminary speed is 0. Any length from aboard the airplane will degree its preliminary speed to be 0, and assuming a reliable point flight, that formula you suggested H = -a million/2*g*t^2 + v0*t + h, is what applies, as measured from the airplane. in the floor reference physique, no, the preliminary speed isn't 0. the project is a 2 dimensional projectile action concern, and it starts off with a HORIZONTAL speed equivalent to that of the airplane. Horizontal place x is: x = v0x*t Vertical postion y is: y = -a million/2*g*t^2 + h For the purpose of this concern, on account that they did no longer inform you v0, you're fixing it in the airplane reference physique, or you're assuming a floor-table sure airplane, like a helicopter soaring at a persevering with top and a persevering with place above the floor.

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- CwCcLv 78 years ago
vf = sqrt((18 m/s)^2 + 2*(3.0 m/s^2)*(50 m)) = 25. m/s

25. m/s [north]

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