need more detail please becuase with this information u only have u and s value does it say anything about dropped from height above ground??? if so we now have a and can use v^2=u^2+2as sorry hope that helps please give full question.
How would I go about answering this question I have no idea how to answer this.
A particle moving with speed vms-1 in direction d has velocity vector v . Find v when v=10 and d= 3i-4j
What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?
What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?
Does it travel that direction vector every second ?
Does it travel that direction vector every second ?
Well no, almost. The direction vector has magnitude 5m. But it travels 10 metres every second. (it must travel at a multiple of the direction vector, every second, but what is that multiple?)
What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?
Actually as it is 5m it travels two direction vectors evey second ?
But how does that then equal a velocity vector :s soo confusssed
The velocity vector is made up of a unit vector multiplied by a magnitude (the speed). The unit vector in the direction of the velocity is given by d divided by its magnitude.
The velocity vector is made up of a unit vector multiplied by a magnitude (the speed). The unit vector in the direction of the velocity is given by d divided by its magnitude.
How would I go about answering this question I have no idea how to answer this. A particle moving with speed vms-1 in direction d has velocity vector v . Find v when v=10 and d= 3i-4j
Velocity =(speed/magnitude of direction vector)×direction vector