Question

a car starting fom rest accelerates at the rate f through a distance s then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest.if the total distance transversed is 15s then find s

Answer 1

Answer on Question #42099, Physics, Mechanics | Kinematics | Dynamics

A car starting from rest accelerates at the rate $f$ through a distance $s$ then continues at constant speed for time $t$ and then decelerates at the rate $f/2$ to come to rest. If the total distance traversed is 15s then find $s$ .

Solution:

The distance at first segment is

$$S _ {1} = S = \frac {v ^ {2}}{2 a _ {1}} = \frac {v ^ {2}}{2 f}$$

The distance at second segment is

$$S _ {2} = \mathrm {B C} = v t$$

The distance at third segment is

$$S _ {3} = \mathrm {C D} = \frac {v ^ {2}}{2 a _ {2}} = \frac {v ^ {2}}{f} = 2 S$$

Thus,

$$S _ {2} = 1 5 S - [ \mathrm {A B} + \mathrm {C D} ] = 1 5 S - S - 2 S = 1 2 S$$ $$S = \frac {v t}{1 2} = \frac {v ^ {2}}{2 f}$$

So,

$$v = \frac {2 f t}{1 2} = \frac {f t}{6}$$

Thus,

$$S = \frac {f t}{6} \frac {t}{1 2} = \frac {f}{7 2} t ^ {2}$$

Answer. $S = \frac{f}{72} t^2$

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