Question

The position of a particle moving in a straight line is described by the relation, x= 5 + 16 t- 4t². Here x is in meters and t in minutes. The distance covered by particle in first 5 minute is​

Answer 1

Answer:

26 m

s=6+12t−2t

2

differentiating with respect to t

v=12−4t

direction is reversed v=0

At t=0,

The displacement is 6 meters

displacement in 3 secs is 6+12

times3−2

times3

2

=24

displacement in 5 secs is 6+12

times5−2

times5

2

=16

Distance travelled in forward direction in 3 sec is 24-6=18 m

Distance travelled in reverse direction in next 2 sec is 24-16=8 m

Thus total distance travelled in 5 sec= 18+8=26 m

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