How Do You Calculate Minimum Constant Acceleration for a Jumbo Jet Takeoff?

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To calculate the minimum constant acceleration for a jumbo jet takeoff, the required speed is 240 km/hr over a runway length of 1.34 km. The formula used is a = vf^2 / (2 * Δx), resulting in an acceleration of approximately 21492.54 km/hr². The discussion highlights the need to convert this value into km/hr/s for clarity. The final solution was confirmed to be correct, emphasizing the importance of unit conversion and significant figures in calculations. The problem was ultimately resolved by the original poster.
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I'm confuse by what the question is asking as far as the conversion (km/hr/s). Can any please help me?

Homework Statement



A jumbo jet must reach a speed of 240 km/hr on the runway for takeoff. What is the minimum constant acceleration (in km/hr/s) needed for takeoff from a 1.34 km runway? (3 sig fig)

Homework Equations





The Attempt at a Solution



a=vf^2/2(Δx)

a=(240 km/hr)^2/2(1.34 km) = (57600 km^2/hr^2) / (2.68 km) = 21492.53731
 
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Ignore this question. Finally solved it!
 
Physicsnoob90 said:
I'm confuse by what the question is asking as far as the conversion (km/hr/s). Can any please help me?

Homework Statement



A jumbo jet must reach a speed of 240 km/hr on the runway for takeoff. What is the minimum constant acceleration (in km/hr/s) needed for takeoff from a 1.34 km runway? (3 sig fig)

Homework Equations


The Attempt at a Solution



a=vf^2/2(Δx)

a=(240 km/hr)^2/2(1.34 km) = (57600 km^2/hr^2) / (2.68 km) = 21492.53731

Your answer is correct, in ##km/hr^2##!...you want to convert units?...

$$21492.53731\frac{km}{hr^2}=21492.53731\frac{km}{hr^2}*\frac{1 hr}{60 min}*\frac{1 min}{60 sec}=? \frac{km}{hr.sec}$$

Remember only three significant figures.

Edit : I guess you solved it!
 

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