Question 59: Motion in 1D

Question

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 :

Accepted Answer

$${t_1} + {t_2} = t,$$ $$V' = 0 + \alpha {t_1}$$
$$V = u + at$$
$$0 = \alpha {t_1} - \beta {t_2}$$
$${t_2} = {\alpha  \over \beta }{t_1} = t$$
$${t_1} = \left( {{\beta  \over {\alpha  + \beta }}} ight)t$$
Distance $$ = {1 \over 2}({t_1} + {t_2}) 	imes \alpha {t_1}$$ (area of triangle)
$$ = {1 \over 2}t 	imes \alpha \left( {{\beta  \over {\alpha  + \beta }}} ight)t$$
$$ = {{\alpha \beta } \over {2(\alpha  + \beta )}}{t^2}$$

Answer

$${{4\alpha \beta } \over {(\alpha  + \beta )}}{t^2}$$

Answer

$${{2\alpha \beta } \over {(\alpha  + \beta )}}{t^2}$$

Answer

$${{\alpha \beta } \over {2(\alpha  + \beta )}}{t^2}$$

Answer

$${{\alpha \beta } \over {4(\alpha  + \beta )}}{t^2}$$

Motion in 1D

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 :

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