An airplane lands on a runway with a velocity of 150 m/s. How far will it travel until it stops if its rate of deceleration is constant -3m/s2?

3750 m

Step-by-step explanation:

This question can be solved using law of equation of motion

where v is the final velocity

u is the initial velocity

a is the acceleration

and

s is the distance travelled

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given

final velocity = v = 0 since it is given that plane stops

u = 150 m/s

a = -3m/s2

using the above values in the given equation

Thus, plane will travel 3750 m before it stops.

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

can i see the following ?

step-by-step explanation: