For a point starting at a point P and travelling along a straight lin

1.	For a point starting at a point P and travelling along a straight line, time of travel is taken as t and the distance from P as s.The relation between s and t is found to be s = 12t – 2t2, where distances to the right are taken as positive numbers and to the left as negative numbers.i. Is the position of the point to the right or left of P, till 6 seconds?ii. Where is the position at 6 seconds?iii. After 6 seconds?(Here it is convenient to write 12t – 2t2 = 2t(6 – t).
Answer» i. S = 12t – 2t² Distance to the point when time is 1 second = 12t – 2t² = 12 × 1 – 2 × 1² = 12 – 2 = 10 m Distance to the point when time is 5 second = 12t – 2t² = 12 × 5 – 2 × 5² = 60 – 50 = 10 m Since the distance to the point till 6 seconds is positive. So the position of the point is on the right of P. ii. Distance to the point when time is 6 second = 12t – 2t² = 12 × 6 – 2 × 62 = 72 – 72 = 0 At the 6^th second, the point is at P. iii. Distance to the point when time is 7 second (after 6 sec)= 12t – 2t² = 12 × 7 – 2 × 7² = 84 – 2 × 49 = 84 – 98 = -14 metres This is a negative number, So the position of the point is on the left of P.

For a point starting at a point P and travelling along a straight line, time of travel is taken as t and the distance from P as s.The relation between s and t is found to be s = 12t – 2t2, where distances to the right are taken as positive numbers and to the left as negative numbers.i. Is the position of the point to the right or left of P, till 6 seconds?ii. Where is the position at 6 seconds?iii. After 6 seconds?(Here it is convenient to write 12t – 2t2 = 2t(6 – t).

Answer»

i. S = 12t – 2t²

Distance to the point when time is 1 second

= 12t – 2t² = 12 × 1 – 2 × 1² = 12 – 2 = 10 m

Distance to the point when time is 5 second

= 12t – 2t² = 12 × 5 – 2 × 5² = 60 – 50 = 10 m

Since the distance to the point till 6 seconds is positive. So the position of the point is on the right of P.

ii. Distance to the point when time is 6

second = 12t – 2t² = 12 × 6 – 2 × 62 = 72 – 72 = 0

At the 6^th second, the point is at P.

iii. Distance to the point when time is 7

second (after 6 sec)= 12t – 2t² = 12 × 7 – 2 × 7²

= 84 – 2 × 49 = 84 – 98 = -14 metres

This is a negative number, So the position of the point is on the left of P.