1.Starting from a point you run 200m east(+ ve x direction) at an average speed

Question

1.Starting from a point you run 200m east(+ve x direction) at an average speed

of 5m/s and then run 280m west at an average speed of 4/s to a post. Calculate

(a)Your average speed from the point to the post

(b) Your average velocity from the point to the post.

2.A car is stopped at a traffic light. It then travels  along a straight road so that its

distance from the  light is given by x(t)=bt2-ct3 where b=2.4m/s2,c=0.120 m/s3.find

(a)Average velocity of the car in the time interval from t=0 to t=10s.

(b)Instantaneous velocity of the car at t=0,t=5,t=10s.

(c ) how long after starting from rest is the car again at rest.

3.A car velocity as a function of time is given by vx(t)= a +bt2 where a =3m/s,

b = 0.1m/s3.find

*(use graph paper or printout from graphic calculator)

4.A pin was thrown upward with an initial speed of 8.2m/s. How much time it took

to return to the ground.

Explanation / Answer

1) You first run 200 m east at 5m/s so it takes you 200/5 = 40 seconds

Then you run 280 m west at 4 m/s so it takes you 280/4 = 70 seconds
So your trip took 110 seconds.

In total you ran 480 meters in 110 seconds
So your speed is 480/110 = 4.3636 m/s

For velocity, you must look at displacement.
You went 200 - 280 m = -80 meters (on the x axis), so 80 meters west
so your velocity is 80/110 meters/s west
which is 0.7272 m/s west.

2)

a) average velocity = total distance / total time

Total distance = 2.4*100 - .12 * 10^3 = 240 - 120 = 120
Average velocity = 12 m/s

b)

Velocity at any instant = dx/dt = 2bt - 3ct^2
At 0 seconds velocity = 0
At 5 seconds velocuty = 2*2.4*5 - 3*.12*25 = 15m/sec
At 10 seconds velocity = 2*2.4* 10 - 3* .12* 100 = -12 m/sec

c)

Velocity is zero when 4.8 t - .36 t^2 = 0
t(4.8-.36t) = 0

either t = 0 which is the start or 4.8-.36t = 0 or t = 13 and 1/3 seconds
approximately 13.33 seconds

3) For this problem some calculus is involved and to do this problem you will need to find the derivative of the equation:

original:

vx(t) = a + bt^2

derivative:

vx'(t) = 2bt

Using the new equation you can find the instantaneous accelereation at t = 5:

b = .0100

vx'(5) = 2b(5)

vx'(5) = 2(.0100)(5)

vx'(5) = .100 m/s^2

4) Tme the pin takes to return to ground = u/g + u/g = 2u/g = 1.673 seconds

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