A) Mind you before your reaction time, you had be going at a uniform speed 18m/s, so for the reaction time of 0.5 seconds, you had covered a distance of:
18m/s*0.5s = 9 m
For the second part which involved deceleration, using:
v = u - at, Noting that there is deceleration.
u = 18m/s, v = final velocity = 0, a = -12m/s².
Let us solve for the time.
<span>v = u + at
</span>
0 = 18 - 12*t
12t = 18
t = 18/12 = 1.5 seconds.
Let us compute for the distance covered during the 1.5s
s = ut + 1/2at², a = -12 m/s²
s = 18*1.5 -0.5*12*1.5² = 13.5m
So the total distance covered = Distance covered from reaction time + Distance covered from deceleration
= 9m + 13.5m = 22.5m
So you have covered 22.5m out of the initial 39m.
Distance between you and the dear: 39 - 22.5 = 6.5m
So you have 6.5m between you and the deer. So you did not hit the deer.
b) Maximum speed you still have:
Well through trial and error, if you maintain the same values of deceleration, reaction time, distance between the car and the deer, you could have a speed of 25 m/s and still not hit the deer. Once it is higher than that by a significant amount you would hit the deer.
Answer:
object with larger mass
Explanation:
the momentum of an object is directly proportional to its mass. larger mass = more momentum
-- If you're observing the moon from a location in the northern hemisphere, like Canada, Russia, the US, or Israel, then it looks like this picture during the <em>waxing crescent</em> phase.
-- If you're observing the moon from a location in the <u>southern</u> hemisphere, like Australia, South Africa, Paraguay, or Antarctica, then it looks like this picture during the <u>waning</u> crescent phase.
The runner's acceleration during this time interval is 10
<u>Given the following data:</u>
- Initial velocity, U = 0 m/s (since the sprinter is starting from rest).
- Final velocity, V = 10.0 m/s
To calculate the runner's acceleration during this time interval, we would use the first equation of motion;
Mathematically, the first equation of motion is calculated by using the formula;
<u>Where:</u>
- U is the initial velocity.
- t is the time measured in seconds.
Substituting the given parameters into the formula, we have;
Therefore, the runner's acceleration during this time interval is 10
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