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BTEC level 3 Mechanical Engineering

I would like assistance on the following question. A component accelerates from rest until it reaches the bottom of the conveyor 4.5 seconds later. The component is travelling at a velocity of 1.35ms when it reaches the bottom of the conveyor where it is then stoppped by hitting a buffer. a) determine the linear acceleration of the component as it travels down the ghravity conveyor. b) The conveyor section length is later modified and extended by a further 1.5m. Determine the new velocity of the components when they reach the buffer.
Reply 1
Original post by fosjoh3
I would like assistance on the following question. A component accelerates from rest until it reaches the bottom of the conveyor 4.5 seconds later. The component is travelling at a velocity of 1.35ms when it reaches the bottom of the conveyor where it is then stoppped by hitting a buffer. a) determine the linear acceleration of the component as it travels down the ghravity conveyor. b) The conveyor section length is later modified and extended by a further 1.5m. Determine the new velocity of the components when they reach the buffer.


SUVAT

V=U+AT

V=1.35
U=0
A=?
T=4.5

1.35 = 0 + 4.5A
A=1.35/4.5
A=0.3 ms-2
Reply 2
Thanks for the help. Do you know the equation to part b?
Reply 3
Original post by fosjoh3
I would like assistance on the following question. A component accelerates from rest until it reaches the bottom of the conveyor 4.5 seconds later. The component is travelling at a velocity of 1.35ms when it reaches the bottom of the conveyor where it is then stoppped by hitting a buffer. a) determine the linear acceleration of the component as it travels down the ghravity conveyor. b) The conveyor section length is later modified and extended by a further 1.5m. Determine the new velocity of the components when they reach the buffer.


Part B

SUVAT to find length of original conveyor -
s=0.5(u+v)t

s=0.5(0+1.35)*4.5
s=3.0375

now
s= 3.0375 + 1.5
s= 4.5375
u=0
v=?
a=0.3

v^2=u^2 + 2as
v^2 = 0 + (2*0.3*4.5375)
v=1.65 ms-1
Reply 4
Part c requests to explain the cause of the components accelerating from rest down the conveyor.
Reply 5
I would very much appreciate some help regarding the following question:

4a) Two vehicles travelling on a road in the same direction collide. Vehicle 1 has a mass 1200kg and is travelling at 30m/s. Vehicle 2 has a mass of 900kg and is travelling at 45m/s. Assuming that the cars momentarily lock together and continue moving in the same direction determine the new velocity (combined) of the cars after the collision.

b) Determine the change in momentum of each vehicle

c) If the time of deceleration of vehicle 2 during the collision is 0.15 seconds determine the average force acting on this vehicle during this time.

d) Determine the avergae forces acting on vehicle 1 during the 0.15 seconds
Reply 6
Original post by fosjoh3
I would very much appreciate some help regarding the following question:

4a) Two vehicles travelling on a road in the same direction collide. Vehicle 1 has a mass 1200kg and is travelling at 30m/s. Vehicle 2 has a mass of 900kg and is travelling at 45m/s. Assuming that the cars momentarily lock together and continue moving in the same direction determine the new velocity (combined) of the cars after the collision.

b) Determine the change in momentum of each vehicle

c) If the time of deceleration of vehicle 2 during the collision is 0.15 seconds determine the average force acting on this vehicle during this time.

d) Determine the avergae forces acting on vehicle 1 during the 0.15 seconds


This should be in the maths thread also don't ask for answers try it yourself first and post your working then wen ur stuck someone in here will help you...

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