Question

A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.8 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 25.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis.

Answer 1

The x component of the dog's velocity is 7.5 m/s at t = 25 seconds.

Given in the question,

Initial Velocity Vx = 2.8 m/s

Initial Velocity Vy = -1.8 m/s

Acceleration = 0.45 m/s2

The direction of acceleration a = (Ф) = 36.0 Degrees

Now acceleration In the X-axis direction,

Ax = Acos(Ф)

Put in the value, we get

Ax = 0.45 × Cos(36)

Ax = 0.45 × 0.809    [ Value of Cos 36 = 0.809 ]

Ax = 0.364 m/s²

Now for X- the axis

Initial Velocity = u = 2.8 m/s

Acceleration = Ax = 0.364 m/s²

Time 1 = t1 = 12 s

Time 2 = t2 = 25 s

Time Interval = t2 - t1

Time Interval = 25 - 12

Time Interval = t = 13 seconds

Now using the first equation of motion,

v = u + at, Put in the value, we get

v = 2.8 + (0.364)(13)

v = 2.8 + 4.732

v = 7.532 m/s

v = 7.5 m/s [ Rounding off to one decimal place]

So at t2 = 25 s, the x component of the dog's velocity is 7.5 m/s.

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A dog running in an open field has components of velocity Vx = 2.8 m/s and vy = -1.8 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 25.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis. At time t2 =25 s, what is the x component of the dog's velocity?