Question

A train starts from rest and acquires a speed $$v$$ with a uniformed acceleration $$\alpha$$. It then comes to a stop with a uniform deceleration $$\beta$$. What is the average velocity of the train?

A
αβα+β
B
α+βαβ
C
v2
D
v

Solution

The correct option is D $$\frac{v}{2}$$From A to B$$S_1= \dfrac{v^2}{2 \alpha}$$           $$t_1= \dfrac{v}{\alpha}$$From B to C$$S_2 = \dfrac{v^2}{2 \beta}$$            $$t_2= \dfrac{v}{\beta}$$avg. velocity $$= \dfrac{S_1 + S_2}{t_1 + t_2}$$                                          $$=\dfrac{\left( \dfrac{v^2}{2 \alpha} + \dfrac{v^2}{2 \beta} \right)}{\dfrac{v}{\alpha} + \dfrac{v}{\beta}} = \dfrac{v}{2}$$General Knowledge

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