Question

Two particles start simultaneously from two points. V
and W with velocities' 144 km/hr and 252 km/hr
respectively in the same direction along the line
VW of length 36 m and with accelerations 8 m/s2
and -4 m/s^2? respectively. If initially first particle is
ahead of second particle then their meeting times
in seconds are :​

Answer 1

144×5/18

40m/s

40=0+10t

4sec=t

252×5/18

25.12m/s

25.12=0+10t

2.512 sec=t

4-2.512=1.488sec

Answer 2

Answer:

t= 2 s ,t= 3 s

Explanation:

For V

u = 144 km/hr                    ( 1 km/hr = 5 /18 m/s)

u = 40 m/s

a = 8 m/s²

For W

u = 255 km/hr                    ( 1 km/hr = 5 /18 m/s)

u = 70 m/s

a =  - 4 m/s²

Lets take at distance x they meet each other after time t

For V

We know that

$s=u t+\dfrac{1}{2}a t^2$

$x=40\times t+\dfrac{1}{2}\times 8 \times t^2$

x = 40 t + 4 t²            -----------1

For W

$x+36=70\times t-\dfrac{1}{2}\times 4 \times t^2$

36+x = 70 t - 2 t²        ------------2

From equation 1 and 2

40 t + 4 t² +36 = 70 t - 2 t²  

6 t² +36 - 30 t =0

t² +6 - 5 t =0

So t= 2 s ,t= 3 s