mechanics!

A cyclist travels to work at an average speed of 3m/s and returns home for tea at an average speed of 9m/s. Calculate her average speed for the whole journey. (not 6m/s?)

Describe and explain how you would use a lightgate and a timer to find the average speed at which a trolley passes a point. :confused:

Reply 1

mubinchoudhury

*girlie*

A cyclist travels to work at an average speed of 3m/s and returns home for tea at an average speed of 9m/s. Calculate her average speed for the whole journey. (not 6m/s?)

Describe and explain how you would use a lightgate and a timer to find the average speed at which a trolley passes a point. :confused:

Ok, the average speed is not 6m/s, because the two journeys took different amounts of time due to the change in speed.

Assume the distance was 18m (could be any number but 18 is the LCM). The cyclist will have been travelling at 3m/s for 6s and 9m/s for 2s. Total time is 8s.

Total distance is 36m. Therefore, 36/8 = 4.5m/s. This is the average.

For the second one, I assumeyou mean one light gate. In this case, put a card of known length on the trolley and set the light gate up in such a way that the timer in on only when the light is broken.Pass the trolly through the gate, and thedistance of the card should be divided by the time on the timer to give an average speed. If you have mre light gates and timers, you can be more accurate.

Reply 2

Jay Mehta

)As a lift moves upwards its motion has three stages

Stage 1 - It accelerates uniformly, from rest, at 0.2ms for 8 seconds.
Stage 2 - it then travels at a contsnat speed for 3 seconds.
Stage 3 - It then decelerates uniformly, stopping when it has travelled a further 1 metre.

ai) Find the distance travelled by the lift in Stage 1.
ii) Find the speed of the lift at the end of Stage 1.
iii) Find the total distance travelled by the lift.

b) Find the total time for which the lift is moving.

c) The lift and its passengers have a total mass of 600kg. The lift is supported by a single cable. Find the tension in the cable when the lift is accelerating upward is 0.2

2) A particle moves on a straight line. At time t seconds its acceleration, a ms is given by

a= 20sin4t

a) Intially the particle is at rest. Find an expresssion for the velocity of the particle at time t.

b) Intially the displacement of the particle from the origin is 0.8meteres. Find an expression for the displacement of the particle at time t.

girlie how are u

Reply 3

SinghFello

2

I remember that question! its in the edexcel work book. It took me ages to get 4.5 too...

Reply 4

RobbieC

15

Stage 1 - It accelerates uniformly, from rest, at 0.2ms for 8 seconds.
Stage 2 - it then travels at a contsnat speed for 3 seconds.
Stage 3 - It then decelerates uniformly, stopping when it has travelled a further 1 metre.

ai) Find the distance travelled by the lift in Stage 1.
s = ut + (1/2)at^2. So s = 0t + (64/2 x 0.2)
s = 6.4m.

ii) Find the speed of the lift at the end of Stage 1.
v^2 = u^2 + 2as. So, v^2 = 0 + 2.56.
Square root of 2.56 = 1.6
v = 1.6

iii) Find the total distance travelled by the lift.
In stage 1 s = 6.4m. In stage 2 s = vt = 1.6 x 3seconds = 4.8m. In stage 3 it travels 1m...
Total s = 12.2m

b) Find the total time for which the lift is moving.
In the first two stages we are told the lift moves for 11 seconds. In the third stage deceleration is constant, so we can calculate the time.
s = ut + (1/2)at^2
1 = 1.6t - a/2(t^2) (multiplying by 10)
10 = 16t - 5at^2 <------ '1'
Finding a: v^2 = u^2 + 2as
0 = 2.56 - 2a
a = 1.28
Putting this into earlier equation '1' gives
10 = 16t - 6.4t^2 ----> 6.4t^2 - 16t + 10
Using the quadratic formula you should now get the answer
t = 1.25s
Therefore total time = 12.25s
t(total) = 12.25s

c) The lift and its passengers have a total mass of 600kg. The lift is supported by a single cable. Find the tension in the cable when the lift is accelerating upward is 0.2
F = ma. Downward force is Weight = mg. Therefore this equals 600g.
We can assume downward and upward forces are equal since the weight of the passengers and the tension are opposite forces (newtons third law of motion).
F = T = 600g x 0.2ms^-2.
T = 120N

I might do 2 later if no one else does... Im just really tired now.. hope that helps you with the first question tho! If you can please rep me, that took ages!

Reply 5

C4>O7

5

*girlie*

A cyclist travels to work at an average speed of 3m/s and returns home for tea at an average speed of 9m/s. Calculate her average speed for the whole journey. (not 6m/s?)

Describe and explain how you would use a lightgate and a timer to find the average speed at which a trolley passes a point. :confused:

The average speed will never be 6ms^-1

http://www.freewebs.com/c4o7/temporary.htm

Reply 6

Fermat

8

Jay Mehta

2) A particle moves on a straight line. At time t seconds its acceleration, a ms is given by

a= 20sin4t

a) Intially the particle is at rest. Find an expresssion for the velocity of the particle at time t.

b) Intially the displacement of the particle from the origin is 0.8meteres. Find an expression for the displacement of the particle at time t.

a= 20sin4t

a)
d²x/dt² = 20sin4t
dx/dt = ∫ 20sin4t dt = -5cos4t + C
at t = 0, dx/dt = 0 (particle is at rest)
=> C = 5
dx/dt = 5(1 - cos4t)
================

b)
dx/dt = 5(1 - cos4t)
x = 5 ∫ 1 - cos4t dt = 5(t - ¼sin4t) + D
at t = 0, x = 0.8
=> D = 0.8 = 4/5
x = 5(4 + t - 0.25sin4t)
==================

mechanics!

Related discussions

Articles for you