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blondinia [14]
3 years ago
9

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acc

eleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above
Physics
1 answer:
Serjik [45]3 years ago
8 0

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the <u>position must also be changing</u>.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the <u>acceleration must be constant</u> too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

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