Question

A body travelling with constant retardation travels 250 metres in 10 seconds and next 250 metres in next 20 seconds. Find the retardation.

Answer 1

A body travelling with constant retardation travels 250 metres in 10 seconds and next 250 metres in next 20 seconds. Find the retardation.

**Solution:**

$$L_1 = L_2 = 250\,m$$ $$t_1 = 10\,s$$ $$t_2 = 20\,s$$

The equation of motion for the first part of the distance (1):

$$x: L_1 = V_1 t_1 - \frac{a t_1^2}{2}$$

Rate equation for the same section of all the way (point A) (1):

$$V_2 = V_1 - a t_1$$ $$V_1 = V_2 + a t_1$$ $$(2)\text{in}(1): \quad L_1 = (V_2 + a t_1) t_1 - \frac{a t_1^2}{2}$$ $$L_1 = V_2 t_1 + \frac{a t_1^2}{2}$$ $$V_2 t_1 = L_1 - \frac{a t_1^2}{2}$$

The equation of motion for the second part of the distance (2):

$$x: L_2 = V_2 t_2 - \frac{a t_2^2}{2}$$ $$V_2 t_2 = L_2 + \frac{a t_2^2}{2}$$ $$(4) \div (3): \quad \frac{V_2 t_1}{V_2 t_2} = \frac{2 L_1 - a t_1^2}{2 L_2 + a t_2^2}$$ $$\frac {t _ {1}}{t _ {2}} = \frac {2 L _ {1} - a t _ {1} ^ {2}}{2 L _ {2} + a t _ {2} ^ {2}}$$ $$2 L _ {2} t _ {1} + a t _ {2} ^ {2} t _ {1} = 2 L _ {1} t _ {2} - a t _ {1} ^ {2} t _ {2}$$ $$a \left(t _ {2} ^ {2} t _ {1} + t _ {1} ^ {2} t _ {2}\right) = 2 \left(L _ {1} t _ {2} - L _ {2} t _ {1}\right)$$ $$a = \frac {2 (L _ {1} t _ {2} - L _ {2} t _ {1})}{t _ {2} ^ {2} t _ {1} - t _ {1} ^ {2} t _ {2}} = \frac {2 (2 5 0 m \cdot 2 0 s - 2 5 0 m \cdot 1 0 s)}{(2 0 s) ^ {2} \cdot 1 0 s - (1 0 s) ^ {2} \cdot 2 0 s} = 2.5 \frac {m}{s ^ {2}}$$

Answer: the retardation is $a = 2.5\frac{m}{s^2}$