Question

The displacement of a particle moving in a straight line is described by the relation, $s=6+12t-2t^2$. Here ${}^{\prime }{s}^{\prime }$ is in metre and ${}^{\prime }{t}^{\prime }$ is in second. The distance covered by particle in first $5sec$ is:

• A. $20m$
• B. $32m$
• C. $24m$
• D. $26m$

Answer 1

The correct option is D $26m$

$s=6+12t-2t^2$

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 $6+12times3-2times{3}^2=24$

displacement in 5 secs is $6+12times5-2times{5}^2=16$

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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