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## Friday, 29 August 2014

## Thursday, 28 August 2014

## Wednesday, 27 August 2014

### On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each

##
On a two-lane road, car A is travelling
with a speed of 36 km h^{-1}. Two cars B and C approach car A in
opposite directions with a speed of 54 km h^{-1} each. At a certain instant,
when the distance AB is equal to AC, both being 1 km, B decides to overtake A
before C does. What minimum acceleration of car B is required to avoid an accident?

## Answer:

Velocity of car A, vA = 36 km/h = 10 m/s

Velocity of car B, vB = 54 km/h = 15 m/s

Velocity of car C, vC = 54 km/h = 15 m/s

Relative velocity of car B with respect to car
A, vBA = vB - vA = 15 - 10 = 5 m/s

Relative velocity of car C with respect to car
A, vCA = vC - (- vA) = 15 + 10 = 25 m/s

At a certain instance,
both cars B and C are at the same distance from car A i.e., s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m =

Hence, to avoid an
accident, car B must cover the same distance in a maximum of 40 s.

### Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

##
Two trains A and B of length 400 m each
are moving on two parallel tracks with a uniform speed of 72 km h^{-1}
in the same direction, with A ahead of B. The driver of B decides to overtake A
and accelerates by 1 m/s^{2}. If after 50 s, the guard of B just brushes
past the driver of A, what was the original distance between them?

Answer:

For train A:

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 s

Time, t = 50 s

Acceleration, aI = 0 (Since it is moving with
a uniform velocity)

From second equation of
motion, distance (sI) covered by train A can be obtained as:

= 20 × 50 + 0 = 1000 m

For train B:

Initial velocity, u = 72 km/h = 20 m/s

Acceleration, a = 1 m/s

Acceleration, a = 1 m/s

^{2}
Time, t = 50 s

From second equation of
motion, distance (sII)covered by train A can be obtained

as:

as:

Hence, the original
distance between the driver of train A and the guard of train B is 2250 -1000 = 1250 m.

### A car moving along a straight highway with a speed of 126 km

##
A car moving along a straight highway
with a speed of 126 km h^{-1} is brought to a stop within a distance of
200 m. What is the retardation of the car (assumed uniform), and how
long does it take for the car to stop?

## Answer:

Initial velocity of the car, u = 126 km/h = 35
m/s Final velocity of the car, v = 0

Distance covered by the car before coming to
rest, s = 200 m Retardation produced in the car = a

From third equation of motion, a can be
calculated as:

From first equation of
motion, time (t) taken by the car to stop can be obtained as:

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