Question

$\huge\bold\pink{Question}$→ An armored car 2m long and 3m wide is moving at 10m/s when a bullet hits it in a direction making an angle tan^-3/4 with the length of the car as seen by a stationary observer.The bullet enters one edge of the car at corner and passes out at the diagonally opposite corner.Neglecting any interaction between the car and the bullet and effect of gravity,the time for the bullet to cross the car is??? Solve krte baitho xD❤~​

Answer 1

• hope it's helpful for you ✌️

Answer 2

Explanation:

Given,

Length, width of car = 2m, 3m

∅ = tan^-1 3/4

tan∅ = 3/4

then, sin∅ = 3/5, cos∅ = 4/5

__________________________________________________

If we separate the components of velocity vector (the path crossed by the bullet while piercing the car) of the bullet along the length and width of the car we get

(vsin∅)t = 3m ...(v = s/t) (1)

Since, the observation is from stationary position the relative velocity between the bullet and the car is

(vcos∅ - 10) (cos component is in the

direction of velocity of the car)

so, (vcos∅ - 10)t = 2m …(2)

putting given values in (1) & (2) we get

$(v( \frac{3}{5} )t )= 3 \\ v \: = \frac{5}{t}$

substituting the value of v in (2)

$(( \frac{5}{t} ) ( \frac{4}{5} )- 10)t = 2 \\ ( \frac{20 - 50t}{5t} )t = 2 \\ \frac{5(4 - 10t)t}{5t} = 2 \\ 4 - 10t = 2 \\ t = \frac{2 - 4}{ -10} \\ t = \frac{1}{5} \\ t = 0.2$

Therefore, the bullet takes 0.2 seconds to cross the car.