Question

A car is crossing point A at time t = 0 with velocity
of 5 ms-1 and reaches to point B with velocity of
5√7 ms'. The distance in between the points A
and B is 100 m. If car is moving with the constant
acceleration, then the time taken by the car to just
cross the mid-point of AB is
(1) 5 s
(3) 7.5 s
(2)20/3 s
(4) 2.5 s​

Answer 1

Answer:

20/3 is the correct answer

Explanation:

AB = 100 m        Let x be the midpoint of the AB  So AX = 50 m

using newton's third law of motion

$\frac{v^2-u^2}{2s} = a$

here, v = $5\sqrt{7}$ m/s     u = 5 m/s            and s = 100 m

put the value and we get a =  0.75 m/s^2

again put this equation of the midpoint

here,    a= 0.75 m/s  ( a = constant)       u = 5m/s         s = 50 m          v =?

put the above values and we get

v = 10 m /s

then using newton's first law

v = u +at

here v = 10m/s     u= 5m/s        a= 0.75 m/s        

we get t = 20/3 sec,

Hope it helps!!

Thanks

Answer 2

Answer:  t = 20/3 sec,

Explanation:AB = 100 m        Let x be the midpoint of the AB  So AX = 50 m

using newton's third law of motion

here, v =  m/s     u = 5 m/s            and s = 100 m

put the value and we get a =  0.75 m/s^2

again put this equation of the midpoint

here,    a= 0.75 m/s  ( a = constant)       u = 5m/s         s = 50 m          v =?

put the above values and we get

v = 10 m /s

then using newton's first law

v = u +at

here v = 10m/s     u= 5m/s        a= 0.75 m/s        

we get t = 20/3 sec,