Question

An airplane must reach at 71ms
from takeoff. If the runway is 1.0km
long. What must the acceleration
be?
(2.52 ms )​

Answer 1

final velocity 'v' = 71m/s

distance 's' = 1000m

intial velocity 'u' =0

therefore,using third equation of motion

v^2- u^2=2*a*s

71*71=2*a*1000

a=71*71/(2*1000

a= 2.52m/s^2

Answer 2

The acceleration of the airplane will be $=2.52m/s^2$

Explanation:

To calculate the acceleration, use third equation of motion.

 $v^2=u^2+2ad$

Here v  is the final speed, u is the initial speed, a is the acceleration and d is the distance traveled.

The airplane reach the speed, v = 71 m/s from takeoff, so u = 0. also given d =1 km.

Substitute the given values in above equation, we get

$(71 m/s)^2=0+2a\times1000m$

$a=\frac{5041}{2000m}$

$a =2.52m/s^2$

Thus, the acceleration of the airplane will be $2.52m/s^2$

#Learn More: equation of motion.

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