A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the dist
1 answer:
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First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed 
, and an accelerated motion on the y-axis, with initial speed 
and acceleration 
:
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
Therefore:
which has two solutions:
is the time of the beginning of the motion,
is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
Therefore:
which has two solutions:
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point: