1) A car travels at a constant speed of 53 km/h as it passes over the road profi

Question

1)

A car travels at a constant speed of 53 km/h as it passes over the road profile shown. The centre of gravity of the car is 0.6m above the road surface. At point A (the bottom of the dip), the radius of curvature of the road is 214 metres, whilst at B (the top of the rise), the radius is 152 metres.
What is the magnitude of the centripital acceleration of the vehicle at point A? (your answer should be in m/s2)

2)

A 1400kg car travels at a constant speed of 65 km/h as it passes over the road profile shown. The centre of gravity of the car is 0.6m above the road surface. At point A (the bottom of the dip), the radius of curvature of the road is 126 metres, whilst at B (the top of the rise), the radius is 254 metres.
What is the magnitude of the Normal force between the vehicle and road at point B? (your answer should be in Newtons)

3)

A 1450kg car brakes from 100km/h to rest in 50 metres at a constant acceleration, determine the total deceleration force applied to the car

Your answer should be quoted in Newtons, but use either the number alone, or N in the answer slot

SO - if you calulate 2000 Newtons, solely insert 2000 or 2000N as your answer

(Note that this question only allows a margin of erro of 2N, so do not round your solution by more than this)

4)

A 1450kg car brakes from 110km/h to rest with a constant braking force of 3500 Newtowns, determine the total distance in metres the car needs to stop.

Your answer should be quoted in metres, but use either the number alone, or m in the answer slot

SO - if you calulate 20 metres, solely insert 20 or 20m as your answer

(Note that this question only allows a margin of error of 2m from the exact value, so do not round allow your solution by more than this)

5)

A 1100 kg car is descending down a slope with the brakes applied to keep the velocity constant. What is the total braking force in Newtons being applied if the angle of the slope is 4.0 degrees?

(acceleration due to gravity is taken as 9.81m/s2, and tolerance from the exact solution is 10 Newtons)

6)

A boy throws a rock from a point A at a height h1= 26.4 meters above the ground towards an obstacle B which is h2=9.1 meters above the ground. With what minimum horizontal velocity 'u' m/s must the rock be thrown in order to have the rock just clear the the obstruction B.Take gravitational acceleration to be 9.81m/s2

HINT-
(i) First of all analyse vertical motion only to find the time 't' (secs) taken for the rock to travel from A to B.
(ii) Now consider the horizontal motion of the rock, and use the time 't' found in (i) above to find the final answer for the horizontal velocity 'u'.

7)

A car travels around a horizontal circular track at a constant speed of 60 km/h. If the radius of the track is 25 metres, what is the magnitude of the cars acceleration in m/s2?

8)

A car travels around a horizontal circular track at an initial speed of 96 km/h. It then accelerates at 5m/s2 for the next 2 seconds (and continues to accelerate at this rate).
If the radius of the track is 39 metres, what is the magnitude of the cars acceleration in m/s2 at the end of the 2 seconds?

Explanation / Answer

1)

53 km/h = 53*1000 / (60*60) m/s = 14.72 m/s

Centripital acceleration = v^2 / R = 14.72^2 / (214 - 0.6) = 1.016 m/s^2

2)

65 km/h = 65*1000/(60*60) m/s = 18.056 m/s

At B, normal force N = mg - mv^2/R

N = 1400*9.81 - 1400*18.056^2 / (254 + 0.6)

N = 11941.36 N

3)

100 km/h = 100*1000/(60*60) m/s = 27.78 m/s

Using v^2 = u^2 - 2as we get

0 = 27.78^2 - 2*a*50

a = 7.716 m/s^2

Force = ma = 1450*7.716 = 11188.27 N

4)

110 km/h = 110*1000/(60*60) m/s = 30.56 m/s

a = F/m = 3500 / 1450 = 2.414 m/s^2

Using v^2 = u^2 - 2as we get

0 = 30.56^2 - 2*2.414*s

s = 193.45 m

5)

Component of force due to weight along the incline = mg*Sin theta = 1100*9.81*Sin 4 = 752.7 N

For constant velocity, Braking force = 752.7 N

6)

For vertical motion, time taken from A to B is given by, h = ut + 1/2*at^2

(26.4 - 9.1) = 0 + 1/2*9.81*t^2

t = 1.878 s

Horizontal distance covered in this time D = u*t

So, u = D/t = D/1.878.................(Since figure is missing, I assume D is to be taken from the figure. It is the horizontal distance between A and B)

7)

60 km/h = 60*1000 / (60*60) m/s = 16.67 m/s

a = v^2 / R

a = 16.67^2 / 25 = 11.11 m/s^2

8)

96 km/h = 96*1000 / (60*60) m/s = 26.67 m/s

Velocity after 2 s is given by v = u + at

v = 26.67 + 5*2 = 36.67 m/s

Centripital acceleration = v^2 / R = 36.67 ^2 / 39 = 34.47 m/s^2

Net acceleration = sqrt (34.47^2 + 5^2) = 34.833 m/s^2

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