Kinematics Question 362
Question: A car accelerates from rest at a constant rate $ \alpha $ for some time, after which it decelerates at a constant rate $ \beta $ and comes to rest. If the total time elapsed it’s t, then the maximum velocity acquired by the car it’s
Options:
A) $ ( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } )t $
B) $ ( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } ) $
C) $ \frac{( {{\alpha }^{{}}}+{{\beta }^{{}}} )t}{\alpha \beta } $
D) $ \frac{\alpha \beta t}{{{\alpha }^{{}}}+{{\beta }^{{}}}} $
Show Answer
Answer:
Correct Answer: D
Solution:
In fig., $ A\text{.}{A _{1}}\text{=}{v _{max\text{.}}}$
=$\alpha {t_1}$=$\beta {t_2} $
$ \text{But t=}{t _{1}}+{t _{2}}=\frac{{V _{\max }}}{\alpha }+\frac{{V _{\max }}}{\beta } $
$ ={v _{\max }}( \frac{1}{\alpha }+\frac{1}{\beta } )={v _{\max }}( \frac{\alpha +\beta }{\alpha \beta } ) $
$ or,{{v} _{\max }}=t( \frac{\alpha \beta }{\alpha +\beta } ) $
BETA
