Question

robbers in a car travelling at 20m/s pass a policeman on a motercycle at rest. The policeman immediately starts chasing the robbers. The policeman accelerates at 3 m/s for 12 s and thereafter travels at a constant velocity. Calculate the distance covered by the policeman before he overtakes the car.​

Answer 1

Robbers in a car travelling at 20m/s pass a policeman on a motercycle at rest. The policeman immediately starts chasing the robbers. The policeman accelerates at 3 m/s for 12 s and thereafter travels at a constant velocity. Calculate the distance covered by the policeman before he overtakes the car.

Answer:-

Here, u=0,t=12 sec , a=3 m/s²

Distance covered by the policeman in 12 seconds is:-

We know that,

s=ut + 1/2at²

→0×12 +1/2(3)(12)²

→1/2(3)(144)

→3×72

→216 m

v = u +at

→0 + 3(12)

→36 m/s

The distance covered by the robber in 12 seconds is :-

Distance = speed × time

→20 m/s × 12 s

→240 m

Suppose the policeman overtakes in 12+t seconds

Then distance = 216 + 36t

Then distance covered by the robber in the same time is

→Speed × time = 20 ( 12 + t )

216 + 36 t = 240 + 20 t

36t-20t = 240-216

16t = 24

t = 24/16

t = 3/2 = 1.5 seconds

the distance covered by the policeman before he overtakes the car in 12 + 1.5 = 13.5 seconds is

216+36t

→216 + 36(1.5)

→216 + 54

→270 m

Hope this helps

Answer 2

Step-by-step explanation:

$\huge\mathrm\star{\boxed{\boxed{Answer:-}}}\star$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\huge\mathrm\star{\boxed{\boxed{Solution:-}}}\star$

$\rightarrow$ $\huge\mathrm\star{\boxed{\boxed{The\ distance\ travelled\ by\ the\ policeman\ is\ 270m}}}\star$

$\Rightarrow$ $\mathrm{\boxed{\boxed{Given:-}}}$

$\Rightarrow$ $\mathrm{u\ =\ 0,\ t\ =\ 12s,\ a\ =\ 3ms²}$

$\Rightarrow$ $\mathrm{Distance\ covered\ by\ police\ in\ 12secs}$

$\Rightarrow$ $\mathrm{s\ =\ ut\ +\ \dfrac{1}{2}at²}$

$\Rightarrow$ $\mathrm{0\ ×\ 12\ +\ \dfrac{1}{2}(3)(12)²}$

$\Rightarrow$ $\mathrm{216m}$

$\Rightarrow$ $\mathrm{v\ =\ u\ +\ at}$

$\Rightarrow$ $\mathrm{36m\s}$

$\Rightarrow$ $\mathrm{Distance\ travelled\ by\ robber\ in\ (12\ +\ t)\ sec}$

$\Rightarrow$ $\mathrm{Diatance\ =\ 216\ +\ 36t}$

$\Rightarrow$ $\mathrm{Speed\ ×\ Time\ =\ 20\ (12\ +\ t)}$

$\Rightarrow$ $\mathrm{2.16\ +\ 36t\ =\ 240\ +\ 20t}$

$\Rightarrow$ $\mathrm{t\ =\ 1.5sec}$

$\Rightarrow$ $\mathrm{Total\ time\ =\ 12\ +\ 1.5sec\ =\ 13.5sec}$

$\Rightarrow$ $\mathrm{Distance\ =\ 216\ +\ 36(1.5)}$

$\Rightarrow$ $\mathrm{216\ +\ 54}$

$\Rightarrow$ $\mathrm{270m}$

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$\Large\mathrm\star{\boxed{\boxed{Hence,\ solved}}}\star$