x = 5t³ + 9t² + 2t +4 m

 

1) Differentiating displacement w.r.t to time, you can obtain the velocity of the car as a function of time.

 

dx/dt = 15t² + 18t + 2 = v

Plugging in t=2.7, v = 159.9

 

2) Same procedure again. Differentiate velocity with respect to time to get the acceleration of the car.

 

dv/dt = 30t + 18

a = 99m/s² at t=2.7s.

 

3) Average acc = change in vel / time = 18.8/15 = 1.25m/s²

 

4) Use equation of motion :  v = u + at

    a = (v - u)/t = 21.1/10 = 2.1m/s²

 

5) Use 2nd equation of motion : s = s₀ + ut + (at²⁾/2

    s = 12(9) + 1.16(81) = 201.96

 

6) 18.8 m/s?? (given)

 

7) Once again use 2nd equation of motion: s = s₀ + ut + (at²⁾/2

    But do take note that, u = +19 since the helicopter is moving up and a = -g since gravity acts downward.

 Plug in the values, you get 234m.

 

8) As I stated in the previous question, the package initially has a velocity 19m/s in the upward direction. Due to the influence of gravity, this velocity decreases, reaches 0 and then becomes negative. The maximum height of the package above the ground is at the instant when its velocity becomes 0.

 

Use v² = u² + 2ah

   0 = 19² + 2(-g)h

 

9) The policeman catches the speeder when the following condition is satisfied:

 

S_s - S_p = 0

where S_s is the displacement of the speeder and S_p is the displacement of the police man.

Use s = ut + 1/2 at²  to calculate their displacements and a time 't'

 

49t - (3.27t²)/2 = 0

Solving,t = 30s.

 

10) The question isnt very clear. Does the rock fall from a height of 65m under the influence of gravity? If so,

 

65 = (1/2)(gt²)

t = 3.6s.

 

The runner can run a distance = 10(3.6) = 35m.

 

I hope I have been helpful. Do contact me if any of the above details are not clear. By the way, I have plugged in the value of g as 10m/s² in all the problems. If you wish to be more accurate, take g as 9.8m/s².

 

Cheers.