Question

train travelling at a speed of 90 km per hour brakes are applied so as to provide a uniform acceleration of 0.5 metre per second square

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Answer 1

Solution :

Speed of train, u = 90 km/h = 90 x 5/18 = 25

m/s

acceleration, a = -0.5 m/s2

Use formula,

V = u + at

Finally train will be rest so, final velocity, v = 0

O = 25 - 0.5t

25 = 0.5t =t = 50 sec

Again, use formula,

S = ut + 1/2at?

Where S is distance travelled before stop

S = 25 x 50 - 1/2 x 0.5 x 502

= 1250 - 1/2 x 0.5 x 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec

Answer 2

Solution :

Speed of train, u = 90 km/h = 90 x 5/18 = 25

m/s

acceleration, a = -0.5 m/s2

Use formula,

V = u + at

Finally train will be rest so, final velocity, v = 0

O = 25 - 0.5t

25 = 0.5t =t = 50 sec

Again, use formula,

S = ut + 1/2at?

Where S is distance travelled before stop

S = 25 x 50 - 1/2 x 0.5 x 502

= 1250 - 1/2 x 0.5 x 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec

Answer 3

Solution :

Speed of train, u = 90 km/h = 90 x 5/18 = 25

m/s

acceleration, a = -0.5 m/s2

Use formula,

V = u + at

Finally train will be rest so, final velocity, v = 0

O = 25 - 0.5t

25 = 0.5t =t = 50 sec

Again, use formula,

S = ut + 1/2at?

Where S is distance travelled before stop

S = 25 x 50 - 1/2 x 0.5 x 502

= 1250 - 1/2 x 0.5 x 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec