Question

53. A car driver travelling with a uniform velocity of 2 m/
s notices a railway crossing at a distance of
435 m from him and also he notices that it is going
to be closed in 10 seconds. First he decides to
cross the level crossing hence a accelerates his
car at the rate of 2 ms-2 for five seconds. Then he
decides to stop the car. So he applies brake and
stops the car exactly before the level crossing
(without following the timer). Calculate the minimum
rate at which he has to decelerate the car so that
he stops the car exactly before the level crossing.
(1) 1.8 m/s2
(2) 18 m/s2
0.18 m/s2
(4) 3.6 m/s2
(
3
)​

Answer 1

Answer

The minimum rate at which he has to decelerate

the car so that he stops the car exactly before

the level crossing = 0.18 m/s^2

EXPLANATION.

For first five seconds

U = 2 m/s

time = 5 seconds

acceleration = 2 m/s^2

From the Newton first equation of kinematics

v = u + at

v = 2 + 2 X 5

v = 12 m/s

From Newton third equation of kinematics

$\large{v}^{2} = {u}^{2} + 2as$

(12) ^2 = (2) ^2 + 2(2) s

144 = 4 + 4s

140 = 4s

s = 35 m

From point B remaining distance

435 - 35 = 400 m

v = 0

u = 12 m/s

Therefore,

From Newton third equation of kinematics

v^2 = u^2 + 2as

(0) ^2 = (12) ^2 + 2 X a X 400

a = 144 / 800

a = 0.18 m/s^2

Answer 2

Answer:

0.18 m/s² will be the correct answer