A car moved 2.00 km towards north, made a perpendicular left turn ran for 4km, made perpendicular right turn moved for 1.00 lcm and stopped. The magnitude of the displacement of the car is
plz tell the procedure of doing the problem also
SreekanthK:
Can u clarify the kilometres that the car moved in each turn?
2 km to north, 4 km perpendicular left turn, 1km perpendicular right turn
Oh ok
Done
Answers
Answered by
2
Distance moved to north=2km
Disctance moved to the left perpendicular=4km
Distance moved to right perpendicular to the before left turn=1km
Magnitude of Displacement = Square root of 3^2+4^2= 25km
= 5km
Disctance moved to the left perpendicular=4km
Distance moved to right perpendicular to the before left turn=1km
Magnitude of Displacement = Square root of 3^2+4^2= 25km
= 5km
Oh
Do u??
I just answered the question. :)
Ok thx:)
Oh yah your thinking was correct
Really
How??
Magnitude of displacement is the sum of the travelled km at the ends. that is 4+1=5
ohhh, i understan now thx alot i will post another question soon
I think my equation was wrong sorry. I corrected it.
Answered by
3
Look at the diagram. Car initially at A. Moved north along AB. Then moved West along BC. Then moved along CD.
The final displacement is AD. Displacement is a vector. It has a magnitude and direction.
Final displacement is THE VECTOR SUM of displacements AB , BC and CD. Take components of displacement along north and along West and add them.
AD² = AE² + DE² = 3² + 4² = 25
AD = 5 km
The final displacement is AD. Displacement is a vector. It has a magnitude and direction.
Final displacement is THE VECTOR SUM of displacements AB , BC and CD. Take components of displacement along north and along West and add them.
AD² = AE² + DE² = 3² + 4² = 25
AD = 5 km
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