Question

The position x of a particle varies with time t as $x=6+12 t-2 t^{2}$ where $x$ is in metre and $t$ in seconds. The distance travelled by the particle in first five seconds is

• A. 16 m
• B. 26 m
• C. 10 m
• D. 36 m

Answer 1

Answer: (B)

Given: $x=6+12 t-2 t^{2}$
At $t=0, x=x_{0}=6\, m$
$t=1 s , x=x_{1}=6+12 \times 1-2 \times 1=16 m$
$t=2 s , x=x_{2}=6+12 \times 2-2 \times 4=22 m$
$t=3 s , x=x_{3}=6+12 \times 3-2 \times 9=24 m$
$t=4 s , x=x_{4}=6+12 \times 4-2 \times 16=22 m$
$t=5 s , x=x_{5}=6+12 \times 5-2 \times 25=16 m$
The distance travelled by the particle in $1^{st }$ second is
$D_{1}=x_{1}-x_{0}=10 m$
The distance travelled by the particle in $2^{\text {nd }}$ second is
$D_{2}=x_{2}-x_{1}=6 m $
Similarly, $ D_{3}=2 m , D_{4}=2 m , D_{5}=6 m$
The distance travelled by the particle in first five seconds is
$D =D_{1}+D_{2}+D_{3}+D_{4}+D_{5}$
$=10 m +6 m +2 m +2 m +6 m =26 m$