A car initially traveling at 21.4 m/s accelerates at a rate of 4.4 m/s2 for 7.5 seconds. What is the final velocity of the car?
1 answer:
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Answer:
A.) The arrow`s range is 624,996 m
B.) The arrow`s range is 846.887 m, when the horse is galloping
Explanation:
We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.
By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.
<u>Equations</u>
X axis:
Y axis:
A.) First, it is necessary to know t, total time.
To figure out t value, we use UAM, since time is determined by this movement.
Now, at the end of the movement, , then
Caculate the segcond degree equation to obtain the two possible values for t:
But, in physics, time it could not be negative, so we take
Caculate now:
B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.
Using the same procedure that item A, caculate X
First, we need to know the new time
And we obtain:
One more time, we take the positive time:
Finally: