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If we consider a ball moving at an acceleration of 5m/s^2$5ms^{-2}$, over a time of 4 seconds, the distance covered by the ball in the first second is 5m$5m$. In the 2nd second will 5 + 5 = 10m$5 + 5 = 10m$. In the third second it will cover a distance of 5 + 5 + 5 = 15m$5 + 5 + 5 = 15m$ and so on and so forth. Now, when we substitute this answer in the equations of motion derived from the area under velocity-time and distance-time graphs, we see a variation:

s = 1/2at^2 = 1/2*5*4^2 = 1/2*5*16$$s = 1/2at^2$$ $$ = 1/2*5*4^2$$ $$ = 1/2*5*16$$ = 40 metres is the distance covered. Now if we go back to our initial description of acceleration we that in the 1st sec = 5m 2nd sec = 10 m 3rd sec = 15 sec 4 sec = 20 sec. Total distance covered in this case is 5 + 10 + 15 + 20 = 50 metres?

40 != 50? Why this disparity between the values? Can someone please explain?!

If we consider a ball moving at an acceleration of 5m/s^2, over a time of 4 seconds, the distance covered by the ball in the first second is 5m. In the 2nd second will 5 + 5 = 10m. In the third second it will cover a distance of 5 + 5 + 5 = 15m and so on and so forth. Now, when we substitute this answer in the equations of motion derived from the area under velocity-time and distance-time graphs, we see a variation:

s = 1/2at^2 = 1/2*5*4^2 = 1/2*5*16 = 40 metres is the distance covered. Now if we go back to our initial description of acceleration we that in the 1st sec = 5m 2nd sec = 10 m 3rd sec = 15 sec 4 sec = 20 sec. Total distance covered in this case is 5 + 10 + 15 + 20 = 50 metres?

40 != 50? Why this disparity between the values? Can someone please explain?!

If we consider a ball moving at an acceleration of $5ms^{-2}$, over a time of 4 seconds, the distance covered by the ball in the first second is $5m$. In the 2nd second will $5 + 5 = 10m$. In the third second it will cover a distance of $5 + 5 + 5 = 15m$ and so on and so forth. Now, when we substitute this answer in the equations of motion derived from the area under velocity-time and distance-time graphs, we see a variation:

$$s = 1/2at^2$$ $$ = 1/2*5*4^2$$ $$ = 1/2*5*16$$ = 40 metres is the distance covered. Now if we go back to our initial description of acceleration we that in the 1st sec = 5m 2nd sec = 10 m 3rd sec = 15 sec 4 sec = 20 sec. Total distance covered in this case is 5 + 10 + 15 + 20 = 50 metres?

40 != 50? Why this disparity between the values? Can someone please explain?!

2 It seems the homework tag applies even if it is not actual homework.
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Acceleration: Value Disparity?

If we consider a ball moving at an acceleration of 5m/s^2, over a time of 4 seconds, the distance covered by the ball in the first second is 5m. In the 2nd second will 5 + 5 = 10m. In the third second it will cover a distance of 5 + 5 + 5 = 15m and so on and so forth. Now, when we substitute this answer in the equations of motion derived from the area under velocity-time and distance-time graphs, we see a variation:

s = 1/2at^2 = 1/2*5*4^2 = 1/2*5*16 = 40 metres is the distance covered. Now if we go back to our initial description of acceleration we that in the 1st sec = 5m 2nd sec = 10 m 3rd sec = 15 sec 4 sec = 20 sec. Total distance covered in this case is 5 + 10 + 15 + 20 = 50 metres?

40 != 50? Why this disparity between the values? Can someone please explain?!