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
Answer:
The plane travels 1650m.
Explanation:
Information Given:
[tex]v_{f} =v_{0} +a*t\\75m/s=145m/s+a*15s\\a=(75m/s-145m/s)/15s\\a=-4.66m/s^{2}[/tex]
Note that acceleration is negative, because the plane is in constant desacceleration.
To know how far does the plane fly, we use following formula:
[tex]s=v_{0} *t+1/2*a*t^{2} \\s=145m/s*15s+1/2*-4.66m/s^{2} *225s^{2} \\s=1650 m[/tex]
