Question

I pull the throttle in my racing plane at a = 12.0 m/s2. i was originally flying at v = 100. m/s. where am i when t = 2.0, t = 5.0, and t=10.0?

Answer 1

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 - $\frac{1}{2}$ 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) - ($\frac{1}{2}$ * 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) - ($\frac{1}{2}$ * 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) - ($\frac{1}{2}$ * 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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