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 as the rate f / 2 to come to rest. If the total distance travelled is 15 S, then

Question

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 as the rate f / 2 to come to rest. If the total distance travelled is 15 S, then (A)  S=ft  (B) S = (1/6) ft2 (C) S = (1/72) ft2 (D) S = (1/4) ft2

Tardigrade Admin · Accepted Answer

The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of

O A = f

and slope of

BC = \frac{f}{2}

v = f t_{1} = \frac{f}{2} t_{2}

∴ t_{2} = 2 t_{1}

In graph area of

△ O A D

gives
distances,

S = \frac{1}{2} f t_{1}^{2}

... (i)
Area of rectangle

A BE D

gives distance travelled in time

t

.

S_{2} = (f t_{1}) t

Distance travelled in time

t_{2}

= S_{3} = \frac{1}{2} f_{2} (2 t_{1})^{2}

Thus,

S_{1} + S_{2} + S_{3} = 15 S

S + (f t_{1}) t + f t_{1}^{2} = 15 S

S + (f t_{1}) t + 2 S = 15 S (S = \frac{1}{2} f t_{1}^{2})

(f t_{1}) t = 12 S

... (ii)
From Eqs. (i) and (ii), we have

\frac{12 S}{S} = \frac{( f t _{1} ) t}{\frac{1}{2} ( f t _{1} ) t _{1}}

t_{1} = \frac{t}{6}

From Eq. (i), we get

∴ S = \frac{1}{2} f (t_{1})^{2}

S = \frac{1}{2} f (\frac{t}{6})^{2} = \frac{1}{72} f t^{2}

Q. 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 as the rate $f /2$ to come to rest. If the total distance travelled is $15 S$ , then

$S = f t$

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

$S = \frac{1}{72} f t^{2}$

$S = \frac{1}{4} f t^{2}$

Q. 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 as the rate f/2 to come to rest. If the total distance travelled is 15S, then

S=ft

S=61​ft2

S=721​ft2

S=41​ft2

Q. 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 as the rate $f /2$ to come to rest. If the total distance travelled is $15 S$ , then

$S = f t$

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

$S = \frac{1}{72} f t^{2}$

$S = \frac{1}{4} f t^{2}$