How far does a plane fly in 15 s while its velocity is changing from 145 m/s to 75 m/s at a uniform rate of acceleration?

Physics

2 answers:

kirza4 [7]3 years ago

6 0

Answer:

The plane travels 1650m.

Explanation:

Information Given:

The initial velocity of the plane v = 75 m/s

Duration of change in velocity t = 15s

The final velocity of the plane after 15 s, vf = 75 m/s

$v_{f} =v_{0} +a*t\\75m/s=145m/s+a*15s\\a=(75m/s-145m/s)/15s\\a=-4.66m/s^{2}$

Note that acceleration is negative, because the plane is in constant desacceleration.

To know how far does the plane fly, we use following formula:

$s=v_{0} *t+1/2*a*t^{2} \\s=145m/s*15s+1/2*-4.66m/s^{2} *225s^{2} \\s=1650 m$

MAVERICK [17]3 years ago

4 0

The solution would be like this for this specific problem:

Given:
- The initial velocity of the plane u = 75 m/s
- Duration of change in velocity t = 15s
- The final velocity of the plane after 15 s, v = 145 m/s
- Let its uniform acceleration be ams^−2

Now we know that v = u + at ⇒ .145 = 75 + a × 15

⇒ a = (145) / 15 = 14/3ms^−2

Now distance traversed during 15s is:

s = u × t + 1/2 × a × t^2 = 75 × 15 + ½ × 14/3 × 15^2 = 1650m

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