Question

two cars start together in the same direction from the same places . the first goes at the speed of 10km/hr and the second goes at the speed 8km/hr and increases he speed by 1/2 km each suceeding hours . after how many hours the second car overtake the first car.if both of them go on non stop

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