1) The driving wheel of a bicycle is rotating at N = 195 revs per minute. If the
ID: 1859969 • Letter: 1
Question
1) The driving wheel of a bicycle is rotating at N = 195 revs per minute. If there is no slippage between the wheel and the road, calculate the bicycle speed in km/h if the diameter of the wheel is 680mm.
2) The position (in metres) of a particle is given by the equation.
What is the velocity of the body at a time 4.3 seconds (in m/s)
Remember, if your solution is negative, include this in your answer.
3) The position (in metres) of a particle is given by the equation.
where the angle will be quoted in radians.
What is the velocity of the body at a time 3.1 seconds (in m/s)
Remember, if your solution is negative, include this in your answer.
What is the acceleration of the body at a time 4.1 seconds (in m/s2)
Remember, if your solution is negative, include this in your answer.
What is the velocity at t = 0.8 seconds (in m/s)
6) For the first several seconds of time, a particle accelerates (in m/s2) at a rate
7) If the initial velocity is 1 m/s, how far from the origin will the particle have moved at a time t = 1.1 seconds (in m/s)
A baseball fielder throws the ball with an initial velocity v0 = 15.0 m/s, at an angle 35 degrees to the horizontal. What is the maximum height (in metres) that the ball will reach? (above the dashed horizontal line shown)
8) A baseball fielder throws the ball with an initial velocity v0 = 29.7 m/s, at an angle 19 degrees to the horizontal. What is the total time (in seconds) taken before the ball hits the ground? Note that the ground is 2 metres below the level from which the ball is thrown.
Please give your answer to 2 decimal places.
9) A baseball fielder throws the ball with an initial velocity v0 = 24.3 m/s, at an angle 34 degrees to the horizontal. What is the total horizontal distance (in metres) travelled before the ball hits the ground? Note that the ground is 2 metres below the level from which the ball is thrown.
Please give your answer to 2 decimal places.
10) A boy throws a rock from a point A at a height h1= 22.1 meters above the ground towards an obstacle B which is h2=10.6 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'.
Explanation / Answer
1) V = pi*D*N/60 = 3.14*0.68*195/60 = 6.939 m/s
2)
v = ds/dt = 4t + 4 = 4*4.3 + 4 = 21.2 m/s
3)
s = 3 sin(4t)
v = ds/dt = 3*4 cos(4t)
v(3.1) = 3*4*cos(4*3.1) = 11.834 m/s
4)
s = 4t3+8.1t2-3t+1
v = ds/dt = 4*3*t^2 + 8.1*2*t - 3
a = dv/dt = 4*3*2t + 8.1*2
a(4.1) = 4*3*2*4.1 + 8.1*2 = 114.6 m/s^2
5)
a = -t2+5.1t
a = dv/dt
dv = a dt
v = Integral a dt
v = Integral -t2+5.1t dt
v = -t^3 /3 + 5.1*t^2 /2 + C
At t= 0 it is given that v = 0.
So, C = 0
Thus, v = -t^3 /3 + 5.1*t^2 /2
v(0.8) = -0.8^3 /3 + 5.1*0.8^2 /2
v(0.8) = 1.461333 m/s
6)
a = -t2+5t
v = Integral a dt
v = -t^3 /3 + 5t^2 /2 + C1
At t = 0, we have v = 1
Thus, C1 = 1
Thus, v = -t^3 /3 + 5t^2 /2 + 1
s = Integral v dt
s = -(1/3) t^4 /4 + (5/2) t^3 /3 + t + C2
At t = 0 we have s = 0.
Thus, C2 = 0
Thus, s = -(1/3) t^4 /4 + (5/2) t^3 /3 + t
s(1.1) = -(1/3) 1.1^4 /4 + (5/2) 1.1^3 /3 + 1.1
s(1.1) = 2.08716 m
7)
Vertical component of velocity v_y = 15 Sin35 = 8.6 m/s
Using, v^2 = u^2 - 2gh we get
0 = 8.6^2 - 2*9.81*h
h = 3.773 m
8)
Vertical component of velocity v_y = 29.7 sin19 = 9.669 m/s
Horizontal component of velocity, v_x = 29.7 cos 19 = 28.08 m/s
For vertical direction, using, v = u - gt we get
0 = 9.669 - 9.81*t
t = 0.9856 s
For vertical direction, using, v^2 = u^2 - 2gh we get
0 = 9.669^2 - 2*9.81*h
h = 4.765 m
Total height = 4.765 + 2 = 6.765 m
Time to reach the ground from 6.765 m height is using h = ut + 1/2*gt^2
6.765 = 0 + 1/2*9.81*t^2
t = 1.1744 m/s
Total time = 1.1744 + 0.9856 = 2.16 s
Horizontal distance covered in 2.16 s is = 28.08*2.16 = 60.653 m
9)
Vertical component of velocity v_y = 24.3 sin34 = 13.59 m/s
Horizontal component of velocity, v_x = 24.3 cos 34 = 20.146 m/s
For vertical direction, using, v = u - gt we get
0 = 13.59 - 9.81*t
t = 1.385 s
For vertical direction, using, v^2 = u^2 - 2gh we get
0 = 13.59^2 - 2*9.81*h
h = 9.413 m
Total height = 9.413 + 2 = 11.413 m
Time to reach the ground from 11.413 m height is using h = ut + 1/2*gt^2
11.413 = 0 + 1/2*9.81*t^2
t = 1.5254 m/s
Total time = 1.5254 + 1.385 = 2.91 s
Horizontal distance covered in 2.91 s is = 20.146*2.91 = 58.63 m
10)
Difference in height h = 22.1 - 10.6 = 11.5 m
Using h = ut + 1/2*gt^2
11.5 = 0 + 1/2*9.81*t^2
t = 1.5311 s
Velocity required = D / 1.5311
where d = horizontal distance between A and B which can be obtained from figure which is missing here.
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