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 $t = 0, x = x_0 = 6 \,m$
$t = 1\,s, x = x_1 = 6+12 \times 1 - 2 \times 1 = 16\,m$
$t = 2\,s, x = x_2 = 6+12 \times 2 - 2 \times 4 = 22\,m$
$t = 3\,s, x = x_3 = 6+12 \times 3 - 2 \times 9 = 24\,m$
$t = 4\,s, x = x_4 = 6+12 \times 4 - 2 \times 16 = 22\,m$
$t = 5\,s, x = x_5 = 6+12 \times 5 - 2 \times 25 = 16\,m$
The distance travelled by the particle in first second is
$D_1 = x_1 - x_0 = 10\,m$
The distance travelled by the particle in second 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
$D = D_1 + D_2 +D_3 + D_4 + D_5$
$= 10\,m + 6\,m +2\,m+6\,m = 26\,m$