Question

The distance between two stations. Is 340km .two trains start simultaneously from these stations on parelle tracks to cross each other .the speed of one of them is greater than that of the other by 5km/hr .if the distance between the two trains after 2hours of their start is 30 km , find the speed of each train

Answer 1

Due to time constraints only 3/5 question will be answered...

 

(1)

distance travelled in the first case

d1 = velocity X time = 40 km/hr X (1/12) hr

or

d1 = 3.33 km

 

distance travelled in second case

d2 = 50 km/hr X (1/6) hr

or

d2 = 8.33 km

 

distance travelled in third case

d3 = 25 km/hr X (1/4) hr

or

d3 = 6.25 km

 

so total distance travelled = 3.33 km + 8.33 km + 6.25 km = 17.91 km 

and 

total time taken = (1/12) hr + (1/6) hr + (1/4) hr = 0.5 hr

 

thus, the average speed will be

vav = total distance travelled / total time taken = 17.91 km / 0.5 hr

thus,

vav  = 35.82 km/hr

 

 

(3)

we know that

v2 - u2 = 2as

 

here

v = 0 m/s (as teh bullet stops eventually)

u = 100 m/s

s = 10 cm = 0.1 m

 

now, by rearranging the above equation, we will get the retardation

a = (v2 - u2) / 2s = -u2/2s

or

a = -(100)2 / (2X0.1)

 

thus, the retardation will be

a = - 50000 m/s2

 

 

(4)

we know that

s = ut + (1/2)at2

 

here

u = 0.5 m/s

a = - 0.05 m/s2

 

and

v2 - u2 = 2as

or in this case

s = -u2 / 2a

or as v = 0, we have

s = - (0.5)2 / -(2X0.05)

or

s = 0.25/0.1

thus, s = 2.5m

 

now, by substituting the values in the first equation we get

5 = 0.5t - (0.05/2)t2

or

0.025t2 - 0.5t + 2.5 = 0

 

this is a quadratic equation which can be solved to get the values of t.

so, we get

t =10 seconds

Due to time constraints only 3/5 question will be answered...

 

(1)

distance travelled in the first case

d1 = velocity X time = 40 km/hr X (1/12) hr

or

d1 = 3.33 km

 

distance travelled in second case

d2 = 50 km/hr X (1/6) hr

or

d2 = 8.33 km

 

distance travelled in third case

d3 = 25 km/hr X (1/4) hr

or

d3 = 6.25 km

 

so total distance travelled = 3.33 km + 8.33 km + 6.25 km = 17.91 km 

and 

total time taken = (1/12) hr + (1/6) hr + (1/4) hr = 0.5 hr

 

thus, the average speed will be

vav = total distance travelled / total time taken = 17.91 km / 0.5 hr

thus,

vav  = 35.82 km/hr

 

 

(3)

we know that

v2 - u2 = 2as

 

here

v = 0 m/s (as teh bullet stops eventually)

u = 100 m/s

s = 10 cm = 0.1 m

 

now, by rearranging the above equation, we will get the retardation

a = (v2 - u2) / 2s = -u2/2s

or

a = -(100)2 / (2X0.1)

 

thus, the retardation will be

a = - 50000 m/s2

 

 

(4)

we know that

s = ut + (1/2)at2

 

here

u = 0.5 m/s

a = - 0.05 m/s2

 

and

v2 - u2 = 2as

or in this case

s = -u2 / 2a

or as v = 0, we have

s = - (0.5)2 / -(2X0.05)

or

s = 0.25/0.1

thus, s = 2.5m

 

now, by substituting the values in the first equation we get

5 = 0.5t - (0.05/2)t2

or

0.025t2 - 0.5t + 2.5 = 0

 

this is a quadratic equation which can be solved to get the values of t.

so, we get

t =10 seconds