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The distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.
We have,
Acceleration of plane i.e. a = 12 m/s²
And,
Velocity of plane i.e. v = 100 m/s
And,
t₁ = 2s
t₂ = 5s
t₃ = 10s
So,
Now,
Using the equation of motion,
i.e.
S = vt - at²
Here,
S = Distance,
v = initial velocity,
a = acceleration
t = time taken
Now,
For t₁ = 2s,
Putting values in above equation we get,
S = (100 * 2) - ( * 12 * 2²)
On solving we get,
S = 200 - 24 = 176 m,
So,
Plane will be at 176m distance.
Now,
For t₂ = 5s,
Putting values in above equation we get,
S = (100 * 5) - ( * 12 * 5²)
On solving we get,
S = 500 - 150 = 350 m,
So,
Plane will be at 350m distance.
Now,
For t₃ = 10s,
Putting values in above equation we get,
S = (100 * 10) - ( * 12 * 10²)
On solving we get,
S = 1000 - 600 = 400 m,
So,
Plane will be at 400m distance.
Hence, we can say that the distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.
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