Question

a train travelling at a speed of 90 km per hour brakes are applied so as to produce a uniform acceleration of -4.5 M per second find how far the train will go before it is brought to rest

Answer 1

Heya..!

refer to the image attached.

train travells 69.49 m after applying the brake. and stops after 5.55 secs.

HOPE THIS HELPS YOU.

Answer 2

Answer:-

Given that

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find out

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to rest

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormula

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2as

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2a

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2aOn substituting the known values we get

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2aOn substituting the known values we gets = (02-252)/2(-0.5) meters

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2aOn substituting the known values we gets = (02-252)/2(-0.5) meters= 625 meters

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2aOn substituting the known values we gets = (02-252)/2(-0.5) meters= 625 metersAnswer

Given thatInitial velocity (u) = 90 km/hour = 25 m.s-1Terminal velocity (v) = 0 m.s-1Acceleration (a) = -0.5 m.s-2Find outHow far the train will go before it is brought to restFormulaAs per the third equation of motion, v2-u2=2asTherefore,The distance travelled by train (s) =(v2-u2)/2aOn substituting the known values we gets = (02-252)/2(-0.5) meters= 625 metersAnswerThe train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.