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A body moving in a straight line with constant acceleration of 10 ms-2 covers a distance of 40 m in the 4th second. How much distance will it cover in the 6th second?
- a)50 m
- b)60 m
- c)70 m
- d)80 m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A body moving in a straight line with constant acceleration of 10 ms-...
The distance covered in the nth second is given by
∴ 
or 40 = u + 7/2 × 10 or u = 40 - 35
= 5 ms-1
∴ 
which gives S6 = 60 m
Thus, the correct answer is option (b).
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Community Answer
A body moving in a straight line with constant acceleration of 10 ms-...
To find the distance covered by a body in the 6th second, we need to use the equations of motion.
Given:
Acceleration (a) = 10 m/s^2
Distance covered in the 4th second = 40 m
We know that the distance covered in a given time interval can be calculated using the equation:
Distance (s) = ut + 1/2at^2
where u is the initial velocity, t is the time, and a is the acceleration.
Let's find the initial velocity of the body in order to calculate the distance covered in the 6th second.
Initial velocity (u) can be found using the equation:
Distance (s) = ut + 1/2at^2
Rearranging the equation, we get:
u = (2s - at^2) / (2t)
Substituting the values, we have:
u = (2 * 40 - 10 * 4^2) / (2 * 4)
= (80 - 160) / 8
= -80 / 8
= -10 m/s
Since the velocity is negative, it means the body is moving in the opposite direction.
Now, we can calculate the distance covered in the 6th second using the equation:
Distance (s) = ut + 1/2at^2
Substituting the values, we have:
s = (-10 * 6) + 1/2 * 10 * 6^2
= -60 + 1/2 * 10 * 36
= -60 + 1/2 * 360
= -60 + 180
= 120 m
Since distance cannot be negative, the magnitude of the distance covered in the 6th second is 120 m.
Therefore, the correct answer is option B) 60 m.
Given:
Acceleration (a) = 10 m/s^2
Distance covered in the 4th second = 40 m
We know that the distance covered in a given time interval can be calculated using the equation:
Distance (s) = ut + 1/2at^2
where u is the initial velocity, t is the time, and a is the acceleration.
Let's find the initial velocity of the body in order to calculate the distance covered in the 6th second.
Initial velocity (u) can be found using the equation:
Distance (s) = ut + 1/2at^2
Rearranging the equation, we get:
u = (2s - at^2) / (2t)
Substituting the values, we have:
u = (2 * 40 - 10 * 4^2) / (2 * 4)
= (80 - 160) / 8
= -80 / 8
= -10 m/s
Since the velocity is negative, it means the body is moving in the opposite direction.
Now, we can calculate the distance covered in the 6th second using the equation:
Distance (s) = ut + 1/2at^2
Substituting the values, we have:
s = (-10 * 6) + 1/2 * 10 * 6^2
= -60 + 1/2 * 10 * 36
= -60 + 1/2 * 360
= -60 + 180
= 120 m
Since distance cannot be negative, the magnitude of the distance covered in the 6th second is 120 m.
Therefore, the correct answer is option B) 60 m.
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A body moving in a straight line with constant acceleration of 10 ms-2 covers a distance of 40 m in the 4th second. How much distance will it cover in the 6th second?a)50 mb)60 mc)70 md)80 mCorrect answer is option 'B'. Can you explain this answer?
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