A vehicle travelling with a uniform velocity was given a uniform acceleration, and in the sixth and tenth seconds after the acceleration began it was observed to travel 15 m and 23 m respectively. What is the acceleration and the initial uniform velocity?

Feb 17, 2018

$a = 2 \text{ m/s"^2}$

$u = 4 \text{ m/s}$

Explanation:

I cannot delete my own question but I figured it out. $\therefore$ I shall answer it.

Using the formula below I shall derive two equations and then solve them simultaneously via substitution

$s = u t + \frac{1}{2} a {t}^{2}$

First, the distance travelled in 6 seconds minus the distance travelled in 5 seconds equals the distance travelled in the sixth second

$15 = \left(6 u + \frac{1}{2} a \left({6}^{2}\right)\right) - \left(5 u + \frac{1}{2} a \left({5}^{2}\right)\right) \Rightarrow 15 = u + \frac{11 a}{2}$

$\Rightarrow u = 15 - \frac{11 a}{2}$ (1)

Second, the distance travelled in 10 seconds minus the distance travelled in 9 seconds equals the distance travelled in the tenth second

$23 = \left(10 u + \frac{1}{2} a \left({10}^{2}\right)\right) - \left(9 u + \frac{1}{2} a \left({9}^{2}\right)\right) \Rightarrow 23 = u + \frac{19 a}{2}$ (2)

Substitute (1) into (2)

$23 = 15 - \frac{11 a}{2} + \frac{19 a}{2} \Rightarrow 8 = 4 a \Rightarrow a = 2 \text{ m/s"^2}$

$u = 15 - \frac{11 a}{2} \Rightarrow u = 4 \text{ m/s}$