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2.Motion in Straight Line
hard
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 is $t$, then the maximum velocity acquired by the car is
A$\left( {\frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}} \right)\,t$
B$\left( {\frac{{{\alpha ^2} - {\beta ^2}}}{{\alpha \beta }}} \right)\,t$
C$\frac{{(\alpha + \beta )\,t}}{{\alpha \beta }}$
D$\frac{{\alpha \beta \,t}}{{\alpha + \beta }}$
(IIT-1978) (AIPMT-1994)
Solution
(d) Let the car accelerate at rate $\alpha $ for time ${t_1}$ then maximum velocity attained, $v = 0 + \alpha {t_1} = \alpha {t_1}$
Now, the car decelerates at a rate $\beta $ for time $(t – {t_1})$ and finally comes to rest. Then,
$0 = v – \beta (t – {t_1})$ $⇒$ $0 = \alpha {t_1} – \beta t + \beta {t_1}$
$⇒$ ${t_1} = \frac{\beta }{{\alpha + \beta }}t$
$v = \frac{{\alpha \beta }}{{\alpha + \beta }}t$
Now, the car decelerates at a rate $\beta $ for time $(t – {t_1})$ and finally comes to rest. Then,
$0 = v – \beta (t – {t_1})$ $⇒$ $0 = \alpha {t_1} – \beta t + \beta {t_1}$
$⇒$ ${t_1} = \frac{\beta }{{\alpha + \beta }}t$
$v = \frac{{\alpha \beta }}{{\alpha + \beta }}t$
Standard 11
Physics
