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A car moving along a straight road with speed of 1.44 km h is brought to a stop within a distance of 200 m. How long does it take for the car to stop?
- a)5 s
- b)10 s
- c)15 s
- d)20 s
Correct answer is option 'B'. Can you explain this answer?
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To find the time it takes for the car to stop, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s in this case), u is the initial velocity (144 km/h or 40 m/s), a is the acceleration, and s is the distance traveled (200 m).
Substituting the values into the equation, we have:
0 = (40 m/s)^2 + 2a * 200 m
Simplifying the equation, we get:
0 = 1600 m^2/s^2 + 400a m
Rearranging the equation, we have:
-400a = 1600
Dividing both sides by -400, we get:
a = -4 m/s^2
Now, we can use the equation of motion again to find the time:
v = u + at
Since the final velocity is 0 m/s and the initial velocity is 40 m/s, we can substitute the values into the equation:
0 = 40 m/s + (-4 m/s^2) * t
Rearranging the equation, we have:
-40 m/s = -4 m/s^2 * t
Dividing both sides by -4 m/s^2, we get:
t = 10 s
Therefore, it takes 10 seconds for the car to stop. The correct answer is option B.
v^2 = u^2 + 2as
where v is the final velocity (0 m/s in this case), u is the initial velocity (144 km/h or 40 m/s), a is the acceleration, and s is the distance traveled (200 m).
Substituting the values into the equation, we have:
0 = (40 m/s)^2 + 2a * 200 m
Simplifying the equation, we get:
0 = 1600 m^2/s^2 + 400a m
Rearranging the equation, we have:
-400a = 1600
Dividing both sides by -400, we get:
a = -4 m/s^2
Now, we can use the equation of motion again to find the time:
v = u + at
Since the final velocity is 0 m/s and the initial velocity is 40 m/s, we can substitute the values into the equation:
0 = 40 m/s + (-4 m/s^2) * t
Rearranging the equation, we have:
-40 m/s = -4 m/s^2 * t
Dividing both sides by -4 m/s^2, we get:
t = 10 s
Therefore, it takes 10 seconds for the car to stop. The correct answer is option B.
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