Question

The table below gives the distance, $d$ feet, the cart was from a reference point at $1$ second intervals while it moves from $t=0$ seconds to $t=5$ seconds.

$t$	$0$	$1$	$2$	$3$	$4$	$5$
$d$	$14$	$20$	$26$	$32$	$38$	$44$

Find the correct relationship between $d$ and $t$.

• A. $d=t+14$
• B. $d=6t+8$
• C. $d=6t+14$
• D. $d=14t+6$
• E. $d=34t$

Answer 1

The correct option is C $d=6t+14$

Observe that in the following series, $d$ follows an Arithmetic progression with $t$ being the term number.

Hence $d_n=d_0+(t-1)\left(r\right)$ where $r$ is the common difference.

Now $r=20-14=26-20=32-26=6$ and $d_0=14$.

Therefore $d_n=14+6\left(\right(t+1)-1)$ since at $t=0,d=14$

$\Rightarrow d=14+6t$