Question

The position x of a particle varies with time t as $x=6+12t-2t^2$ 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: $x=6+12t-2t^2$
At $t=0,x=x_0=6m$
$t=1s,x=x_1=6+12\times 1-2\times 1=16m$
$t=2s,x=x_2=6+12\times 2-2\times 4=22m$
$t=3s,x=x_3=6+12\times 3-2\times 9=24m$
$t=4s,x=x_4=6+12\times 4-2\times 16=22m$
$t=5s,x=x_5=6+12\times 5-2\times 25=16m$
The distance travelled by the particle in $1^{st}$ second is
$D_1=x_1-x_0=10m$
The distance travelled by the particle in $2^{\text{nd }}$ second is
$D_2=x_2-x_1=6m$
Similarly, $D_3=2m,D_4=2m,D_5=6m$
The distance travelled by the particle in first five seconds is
$D=D_1+D_2+D_3+D_4+D_5$
$=10m+6m+2m+2m+6m=26m$