Question

a train at rest start moving with uniform acceleration 2 min reach 48kmh after 5 min it uniform retardation and start after 3 min.how much distance it covered

Answer 1

Task:

A train at rest starts moving with uniform acceleration, keeps moving for 2 min. It reaches 48 km/h after 5 min of its uniform retardation and stops after 3 min. How much distance it covered?

Solution:

Given:

$$t _ {a c c} = 2 \min = \frac {1}{3 0} h$$ $$t _ {r e t} = 5 \min = \frac {1}{1 2} h$$ $$t _ {s t o p} = 3 \min = \frac {1}{2 0} h$$ $$\dot {s} _ {2} \left(t _ {r e t}\right) = 4 8 \mathrm {k m} / h$$ $$\dot {s} _ {2} \left(t _ {r e t} + t _ {s t o p}\right) = 0 \mathrm {k m} / h$$ $$s _ {1} (t) = \frac {a _ {1} \cdot t ^ {2}}{2}$$ $$s _ {2} \left(t _ {1}\right) = \frac {a _ {2} \cdot t _ {1} ^ {2}}{2} + \dot {s} _ {1} \left(t _ {a c c}\right) \cdot t _ {1}$$ $$s _ {t o t a l} = s _ {1} \left(t _ {a c c}\right) + s _ {2} \left(t _ {r e t} + t _ {s t o p}\right)$$ $$s _ {1} \left(t _ {a c c}\right) = \frac {a _ {1} \cdot t _ {a c c} ^ {2}}{2}$$ $$\dot {s} _ {1} (t) = a _ {1} \cdot t$$ $$\dot {s} _ {1} \left(t _ {a c c}\right) = a _ {1} \cdot t _ {a c c}$$ $$\dot {s} _ {2} \left(t _ {1}\right) = a _ {2} \cdot t _ {1} + \dot {s} _ {1} \left(t _ {a c c}\right) = a _ {2} \cdot t _ {1} + a _ {1} \cdot t _ {a c c}$$ $$\dot {s} _ {2} \left(t _ {r e t}\right) = a _ {2} \cdot t _ {r e t} + a _ {1} \cdot t _ {a c c} = 4 8 \mathrm {k m} / h$$ $$\dot {s} _ {2} \left(t _ {r e t} + t _ {s t o p}\right) = a _ {2} \cdot \left(t _ {r e t} + t _ {s t o p}\right) + a _ {1} \cdot t _ {a c c} = 0 \mathrm {k m} / h$$ $$a _ {1} = 3 8 4 0 \mathrm {k m} / h ^ {2}, a _ {2} = - 9 6 0 \mathrm {k m} / h ^ {2};$$ $$s _ {t o t a l} = \frac {a _ {1} \cdot t _ {a c c} ^ {2}}{2} + \frac {a _ {2} \cdot (t _ {r e t} + t _ {s t o p}) ^ {2}}{2} + a _ {1} \cdot t _ {a c c} \cdot (t _ {r e t} + t _ {s t o p}) = \frac {3 2}{3} \mathrm {k m}$$

Answer:

The train covered $\frac{32}{3}$ km