Question

A car travel 2m in 2nd second and 6m in next 4 second. what will be distance travel by car in 9th second?

Answer 1

A car travel 2m in 2nd second and 6m in next 4 second. what will be distance travel by car in 9th second?

Suppose, the initial speed of the car was $v_0 \, m/s$. Then the speed at the time instant $t$ equals:

$$v(t) = v_0 + a t$$

a – acceleration

Distance traveled in the $2^{\text{nd}}$ second equals:

$$d_1 = (v_0 + a * 1s) 1s + \frac{a(1s)^2}{2} = v_0 + \frac{3}{2}a$$

$v_0 + a * 1s$ – speed at the beginning of the $2^{\text{nd}}$ second

Distance traveled in the next 4 seconds equals:

$$d_2 = (v_0 + 2s a) 4s + \frac{a(4s)^2}{2} = 4v_0 + 16a$$ $$\left\{ \begin{array}{l} v_0 + \frac{3}{2}a = 2 \\ 4v_0 + 16a = 6 \end{array} \right.$$

(2)-4*(1):

$$4v_0 - 4v_0 + 16a - 6a = 6 - 8$$ $$a = -\frac{1}{5} \, m/s^2 \quad v_0 = 2 - \left(-\frac{1}{5} * \frac{3}{2}\right) = 2 \frac{3 \, m}{10 \, s} = 2.3 \, m/s$$

The distance traveled by car in the 9th second:

$$d = (v_0 + 8s * a) 1s + \frac{a(1s)^2}{2} = 2.3 - \frac{8}{5} - \frac{1}{10} = \frac{6}{10} \, m = 0.6 \, m$$

Answer: 0.6 m