Question

a body travels 200cm in first 2seconds and 220cm in next 5second.Calculate the velocity at the end of the seventh second from the start

Answer 1

a body travels 200cm in first 2seconds and 220cm in next 5second. Calculate the velocity at the end of the seventh second from the start.

Suppose, initial speed of body was $v_0 \, m/s$. Then speed at moment time $t$ equals:

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

a – acceleration

Distance traveled in first 2 second equals:

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

Distance traveled in next 5 second equals:

$$d_2 = (v_0 + 2 s e c * a) 5 s e c + \frac{a (5 s e c)^2}{2} = 5 v_0 + \frac{45}{2} a$$

where $v_0 + 2s e c * a$ velocity at end of $2^{\text{nd}}$ second.

$$\left\{ \begin{array}{l} 2 v_0 + 2 a = 2 \\ 5 v_0 + \frac{45}{2} a = 2.2 \end{array} \right.$$

$2^*(2) - 5^*(1)$:

$$10 v_0 - 10 v_0 + 45 a - 20 a = 4.4 - 20$$ $$a = -\frac{15.6 \, m}{25 \, s^2} = -0.624 \, \frac{m}{s^2}, \quad v_0 = 1 - a = 1.624 \, \frac{m}{s}$$

Velocity at the end of the seventh second equals:

$$v = v_0 + a * 7 = 1.624 - 0.624 * 7 = -2.744 \, \frac{m}{s}$$

Answer: $-2.744 \, \frac{m}{s}$