Physics Q's-need help asap

[ahmed]

Q1
You are to drive to an interview in another town, at a distance of 300km on a freeway. The interview is at 11:15 am. You plan to drive 100km/hr, so you leave at 8 am to allow for some extra time. You drive at that speed for the first 100 km, but then road works force you to slow to 40 km/hr for 40 min. What would be the least speed needed for the rest trip to arrive in time for the interview?

Q2
The position of a particle moving along the x-axis is given in cm by x(t)=9.75 + 1.50t^3, where t is in seconds. Calculate

(a)the average velocity during the time interval t=2sec to t=3sec
(b)the instantaneous velocity at t=2sec
(c)the instantaneous velocity at t=3sec
(d)the instantaneous velocity at t=2.5sec
(e)the instantaneous velocity when the particle is midway between its positions at t=2sec and t=3sec
(f)graph x versus t and indicate results graphically

Q3
The brakes on your car can slow you at a rate of 5.2m/s^2.
(a)If you are going 137 km/hr and suddenly see a speed camera, what is the minimum time in which you an get your car under 90 km/hr speed limit?
(b)Sketch graphs of position versus time and velocity versus time for such a slowing.

Q4
When startled, an armadillo will leap upward. Suppose it rises 0.544 m in the first 0.200 seconds.
(a)what is its initial speed as it leaves the ground?
(b)what is its speed at a height of 0.544 m?
(c)how much higher does it go?

Q5
A ball is shot vertically upward from the surface of a planet in a distant solar system. A plot of y versus t for the ball is shown below, where y is the height of the ball above its starting point and t=0 at the instant the ball is shot.(picture below,PARABOLA)
What are the magnitudes of
(a)free-fall acceleration on the planet and
(b)the initial velocity of the ball?

Q6
A bank in downtown Boston is robbed(see map below).To elude the polce, the robbers escape by helicopter, making three successive flights described by the following 3 displacements:32km, 45 deg. south of east;53 km, 26 deg. north of west;26km, 18 deg. east of south. At the end of the third flight they are captured. In what town are they apprehended?

Q7
A golf ball is struck at ground level. The speed of the golf ball as a function of time is shown below(upside down parabola), where t=0 at the instant the ball is struck.
(a)How far does the golf ball travel horizontally before turning to ground's level?
(b)What is the maximum height above ground level attained by the ball?

Q8
The fast french train TGV has a scheduled av. speed of 216km/hr.
(a)If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is said to be limited to 0.050g, what is the smallest radius of curvature for that track that can be tolerated?
(b)At what speed must the train go around a curve with a 1km radius to be at the acceleration limit?

Thankz for the help in advance.

[nak]

1)
well the first part driving 100km/hr for a distance of 100km means 1hr is lost. so you got 2hrs 15mins and 200kms to go


due to the road works:

v=d/t     so  d =  v * t       d = 40 * (40/60) {you want it in hrs}  = 26.67km

so now you got 173.33kms and 1hr 35mins to go.

1hr 35 mins = 95 mins

so now we need to convert 95mins to hr which is 95/60 = 1.583hrs

so now we need to travel 173.33km in 1.583hrs

then the speed would be v = 173.33/1.583 = 109.5km/hr

Excuse my working out I don't know how to use LaTeX

[kamil9876]

2.)

a.)


b.) Instantaneous velocity is the derivative of position (although calculus is not expected in yr12 physics so don't worry too much.




e.) Let the particle be midway x(3) and x(2)(it's position at t=3 and 2 respectively). Because of this the distance between x(3) and x(t) must equal to the distance between x(t) and x(2).
Now, this function is an always increasing function, so we don't have to worry about turning points.
So:
   (note: because x(2)<x(t)<x(3) be sure to subtract the way i did to get positive values)

Which is basically that midpoint formula u learn in maths.

so now u can use this to find x(t) and so u then plug that in to the original equation to get t, the time when the thing is midway in between. Once u know this t, u can now find v(t) since u already know v(t) from differetiating in part b.

f.) so include y intercept, ensure that u only include non-negative values for time. You may also want to include the tangents whose gradients u worked out in parts b to e. Basically, just include stuff that u worked out in previous parts of the question (the question didn't specify, but in VCEland u can assume these sorts of things from vague and unrigourous questions)

edit: algebraic mistkae in midpoint formula

[MattPritchard]

Q3
The brakes on your car can slow you at a rate of 5.2m/s^2.
(a)If you are going 137 km/hr and suddenly see a speed camera, what is the minimum time in which you an get your car under 90 km/hr speed limit?
(b)Sketch graphs of position versus time and velocity versus time for such a slowing.

Covert -5.2m/s into k/h so units are the same

A m/s x 3.6 = B k/h

-5.2 * 3.6 = 18.72k/h

Use Motion Equation
u=137
v=90
a=-18.72
t=???

v=u+at

Final Speed   =   Starting Speed  +  Rate of Deceleration * Time
     90k/h      =          137k/h      +             -18.72k/h    *   t

(90)=(137)+(-18.72)*(t) (take 137 from both sides)
(90)-(137)=(-18.72)*(t)
(-47)=(-18.72)*(t) (Divide both sides by Negative 18.72)
(-47)/(-18.72)=(t)
(2.5107)=(t)
t=2.51
t=2.51 seconds