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A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled is
- a)32 m
- b)16 m
- c)4 m
- d)8 m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A particle accelerates from rest at a constant rate for some time and...
8 = a t1 and 0 = 8 – a (4 – t1)
8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec
Now, s1 = 0 x 2 + ½ x 4(2)2 or s1 = 8 m
s2 = 8 x 2 - 1/2 x 4 x (2)2 or s2 = 8m
∴ s1 + s2 = 16 m
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Community Answer
A particle accelerates from rest at a constant rate for some time and...
To solve this problem, we can use the equations of motion. Let's break down the problem into two parts: the acceleration phase and the deceleration phase.
Acceleration phase:
Given:
Initial velocity (u) = 0 m/sec
Final velocity (v) = 8 m/sec
Time taken (t) = unknown
We can use the equation of motion:
v = u + at
Substituting the values, we get:
8 = 0 + a*t
Simplifying, we have:
8 = at
Deceleration phase:
Given:
Initial velocity (u) = 8 m/sec
Final velocity (v) = 0 m/sec
Time taken (t) = unknown
Using the same equation of motion:
v = u + at
Substituting the values, we get:
0 = 8 + a*t
Simplifying, we have:
at = -8
Total time:
The total time taken is given as 4 seconds. Therefore, the sum of the time taken during the acceleration and deceleration phases is equal to 4 seconds.
t + t = 4
2t = 4
t = 2 seconds
Distance traveled:
To find the distance traveled, we need to calculate the distance covered during the acceleration phase and the deceleration phase, and then sum them up.
Acceleration phase:
Using the equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s1 = 0*2 + 0.5*a*2^2
s1 = 2a
Deceleration phase:
Using the same equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s2 = 8*2 + 0.5*(-8)*2^2
s2 = 16 - 16
s2 = 0
Therefore, the total distance traveled is the sum of s1 and s2:
s = s1 + s2
s = 2a + 0
s = 2a
Since s = 16 meters (given in option B), we can solve for a:
2a = 16
a = 8
Therefore, the distance traveled is 16 meters, which is given in option B.
Acceleration phase:
Given:
Initial velocity (u) = 0 m/sec
Final velocity (v) = 8 m/sec
Time taken (t) = unknown
We can use the equation of motion:
v = u + at
Substituting the values, we get:
8 = 0 + a*t
Simplifying, we have:
8 = at
Deceleration phase:
Given:
Initial velocity (u) = 8 m/sec
Final velocity (v) = 0 m/sec
Time taken (t) = unknown
Using the same equation of motion:
v = u + at
Substituting the values, we get:
0 = 8 + a*t
Simplifying, we have:
at = -8
Total time:
The total time taken is given as 4 seconds. Therefore, the sum of the time taken during the acceleration and deceleration phases is equal to 4 seconds.
t + t = 4
2t = 4
t = 2 seconds
Distance traveled:
To find the distance traveled, we need to calculate the distance covered during the acceleration phase and the deceleration phase, and then sum them up.
Acceleration phase:
Using the equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s1 = 0*2 + 0.5*a*2^2
s1 = 2a
Deceleration phase:
Using the same equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s2 = 8*2 + 0.5*(-8)*2^2
s2 = 16 - 16
s2 = 0
Therefore, the total distance traveled is the sum of s1 and s2:
s = s1 + s2
s = 2a + 0
s = 2a
Since s = 16 meters (given in option B), we can solve for a:
2a = 16
a = 8
Therefore, the distance traveled is 16 meters, which is given in option B.
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A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer?
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