JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 8)
A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
$${{4\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$
$${{2\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$
$${{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}$$
$${{\alpha \beta } \over {4(\alpha + \beta )}}{t^2}$$
Explanation
_17th_March_Morning_Shift_en_8_1.png)
$${t_1} + {t_2} = t,$$ $$V' = 0 + \alpha {t_1}$$
$$V = u + at$$
$$0 = \alpha {t_1} - \beta {t_2}$$
$${t_2} = {\alpha \over \beta }{t_1} = t$$
$${t_1} = \left( {{\beta \over {\alpha + \beta }}} \right)t$$
Distance $$ = {1 \over 2}({t_1} + {t_2}) \times \alpha {t_1}$$ (area of triangle)
$$ = {1 \over 2}t \times \alpha \left( {{\beta \over {\alpha + \beta }}} \right)t$$
$$ = {{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}$$
Comments (0)
