Kinematics Problems

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

t1 t2 t3t

x

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

• Q. Figure shows x-t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero.

• SOULTION: • We know that average velocity is given by

• According to question, ∆v = 0, therefore,

x2 – x1 = 0

x2 = x1

• Thus, avg. velocity becomes zero when final position = initial position.• From graph, x1 = 20 m at time 0. • Thus, x2 = x1 = 20 m. • From graph we find x2 = 20 m at time 12 s (approximately) . (Ans)

5t (second)

12

x1

10

10

20 ∆v =

∆ x

∆ t =

x2 – x1

t2 – t1

0 =

x2 – x1

t2 – t1

x (m)

x2

Kinematics

• Q A particle starts from point A and travels along the solid curve. Find approximately the position B of the particle such that the avg. velocity between the portions A and B has the same direction as the instantaneous velocity at B.

• Solution:

• We know that instantaneous velocity is tangent to path at a particular instant of time.

• Also, average velocity between a time interval is along the chord joining those two positions.

• Using these two points in mind, we can find the required position B.

• Thus, the required position is x = 5 m & and y =3 m. (Ans)

2x (m)

5

A

4

2

4

y (m)

B3

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

• From the velocity (along x axis) vs. time plot shown, find (a)the distance travelled by the particle during first 40s, (b) the avg. velocity during this period, (c) avg. acceleration between 0 to 10s, and (d) acceleration at t=5s.

• SOLUTION:

t(s)

v (m/s)

0 20 40

5

-5

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

• From theory we know that, from area of velocity time graph we get change in position, i.e., displacement. We also know that area above t-axis is positive and below t-axis is negative.

• Therefore, displacement between t=0s to t=20s = area of v-t graph = area of triangle = ½ x 20 x 5 =50m.• And, displacement between t=20s and t=40s = - (area of v-t graph) [-ve, because area below t-axis] = - ½ x 5 x 20 = -50m.• (a) Distance is the total length of the path. Therefore distance travelled by the particle during the first 40s = 50 + 50 =100m (ans)

(b). Avg. velocity during first 40s = (change in position during this period)/(time change) =(total displacement during this period)/(total time taken) =(50-50)/(40-0) = 0. (ans)

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

(c). Avg. acceleration 0 to 10s = (change in velocity)/(time taken) = (5-0)/(10-0) = 0.5 m/s2. (ans)

(d). We have to find acceleration at t=5s, i.e., we have to find instantaneous acceleration (because at a required acceleration is at particular instant of time) at t=5s.

Instantaneous acceleration is found by slope of tangent at that instant. Therefore, a= tanӨ = p/b =5/10 = 0.5 m/s2. (ans)

t(s)

v (m/s)

0 20 40

5

-5

5

.Ө

10

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

P. K. BHARTI, I.I.T. KHARAGPUR

• A particle moving along the x-axis with a constant acceleration. When t=0, x=4m and v=3m/s. Also, when t=4s, a maximum value of x is obtained. Determine xmax and the value of x when t=12s.

• SOLUTION: As particle is moving along x axis with uniform acceleration such that after t=4s position x is maximum, it means position x starts decreasing after t=4s. It means velocity becomes zero at t=4s.

xOrigint=0sx1 = 4mu = 3m/s

t=4sx2= xmax

v=0

a=?

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

• First of all we have to find a. As we know u, v and t (as t1=0, t=4-0=4s). • We use eqn. v = u + at to find a.• Therefore, v = u + at • 0 = 3 + 4a a=-¾ m/s2 .

• Now, we use eqn x2 – x1 = ut + ½ at2 to find x2=xmax .

• x2 – x1 = ut + ½ at2 xmax –4=3x4 + ½ (-¾)42

• xmax = 10m (ans).

• Similarly we can find x at t=12s• Do it yourself.• Ans: @t=12s, x=-14m.

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

• NOTE : In this question we found acceleration to be negative, which is also obvious from given physical conditions because, particle is initially moving along x-axis but position x increases to maximum and then decrease. It means velocity magnitude is decreasing along positive x axis until x becomes maximum. It means there is a deceleration (-ve acceleration).

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

P. K. BHARTI, I.I.T. KHARAGPUR

• EXAMPLE: A particle starts from origin at t=0s with a velocity of 8j m/s and moves in the xy plane with a constant acceleration of (4i + 2j)m/s2. At the instant the particle’s x coordinate is 29m, what are (a)its y coordinate and(b) its speed?

• SOLUTION: GIVEN QUANTITIES: u = ux i+uy j = 8j m/s ux = 0 and uy = 8m/s.

• a = ax i+ ay j = (4i + 2j)m/s2

• ax = 4m/s2 and ay = 2m/s2.

• Let at a time t=t, x=29m. • Now, We break motion along x axis and y axis.

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

• Along x-axis• x = uxt + ½ axt2 x 29 = 0 + ½ x 4 x t2 • t =√ (29/2) s = 3.8 m/s.

• And, vx = ux + axt = 0 + 4 x 3.8 = 15.2 m/s.

• Along y-axis • y = uyt + ½ ayt2 y = 8x3.8 + ½ x 2 x (√ (29/2))2

• y = 49.96 m = 50m [approx].

• And, vy = uy + ayt = 8 + 2 x 3.8 = 15.6m/s

Therefore, speed v = √(vx2 +vy

2) = 21.78m/s = 22m/s .

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

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• Q. A staircase contains three steps each 10cm high and 20cm wide. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane?

• SOLUTION: • This question is little tricky. • The basic idea is that velocity is minimum for the case

when the ball just crosses (very close but not touching) the last step at point P.

• Let us choose our x-y coordinates as shown in the figure. • Clearly, the coordinates of point P (x , y) is (40 cm, – 20 cm),

i.e., (0.4 m, – 0.2 m). • Let the minimum horizontal velocity be u. • Now, using equation of trajectory, we get

u = 1.97 m• Neglecting negative value, we get u = 1.97 m/s ≈ 2 m/s. (Ans)

Y

x

P

u

y = x tan ө –

g x 2

2u2 cos2 ө

– 0.2 = 0.4 × tan 00 – (because ө = 00 )

9.8 × 0.42

2 × u2 × cos2 00

KINEMATICS

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur

• Q. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with a an acceleration of 1m/s2 and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

• SOLUTION: • Let us visualize the animation.• This problem will be solved beautifully using relative motion concepts.

• BALL: (in the frame of man)• The ball has two velocity components: vertical (because of throw) and horizontal (because of

inertia). • Let us first calculate the time taken by ball to come back to the railroad. It is twice the time taken

by ball to reach the maximum height (neglecting the height of man).

x

y

Kinematics

• Vertical upward motion up to maximum height : Initial vertical component of velocity of ball wrt car uy = 9.8 m/s (because of throw)

Final vertical component of velocity of ball wrt car vy = 0

vertical component of acceleration of ball wrt car ay = –g = –9.8m/s2

• Using vy = uy + ayt in the vertical direction we get,

0 = 9.8 + (– 9.8) t (because, ay = – g = –9.8m/s2)

t = 1s. • Therefore, total time taken by ball to come back to the railroad = 2s.

• Horizontal motion: Initial horizontal component of velocity of ball wrt man ux = 0

(because, ball remains at rest wrt car before throw))horizontal component of acceleration of ball wrt man ax = – (acceleration of car) = – 1 m/s2

• Therefore, using x = ux t + ½ axt2 we get,

x = 0 + ½ (–1) (2) 2 = – 2 m. • Hence, ball falls at distance 2m behind the man.

x

y

Kinematics

• Q. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

• SOLUTION: • The basic idea is that truck is not so large and also 14.7 m/s (52.92 km/h) is a pretty high speed. Thus,

the statement “ ball returns to truck” means that it comes back to person again. • Let us visualize the animation.

• Let us first calculate the time taken by the ball to come back to the truck. It is the time taken by the truck to move a distance 58.8 m at a constant speed of 14.7 m/s.

• Thus, total time taken by ball to come back to the truck = x/v = 58.8/14.7 = 4s.

x

y

Kinematics

• (a) BALL: (in the frame of truck)• It is obvious from animation that ball is thrown vertically upward wrt man (or truck) [because it

comes back to man’s hand again.]

• Vertical upward motion up to maximum height wrt truck : Initial velocity of ball wrt truck uy = u (say)

Final velocity (at maximum height) of ball wrt truck vy = 0

vertical component of acceleration of ball wrt car ay = –g = –9.8m/s2

Time taken by ball to reach the maximum height =4/2 = 2s.

• Using vy = uy + ayt in the vertical direction we get,

0 = u + (– 9.8) 2 (because, ay = – g = –9.8m/s2)

u = 19.6 m/s.

• Therefore, projection speed of the ball as seen from the truck = 19.6 m/s (vertically upward) (Ans)

x

y

Kinematics

• (b) BALL: (in the frame of road)• It is obvious from animation that a man from the road sees the ball moves horizontally as well as

vertically. Thus in the frame of road ball has two components of projection speed. • The ball will have horizontal component of velocity same as that of the truck in this frame.• Thus, horizontal component of projection velocity of the ball wrt ground ux=14.8 m/s. • Also, vertical component of projection velocity of the ball wrt ground = projection speed of the ball as seen from the truck = 19.6 m/s uy= 19.6 m/s

• Hence, projection speed of the ball wrt ground • u = √( ux

2 + uy2 ) = √( 14.82 + 19.62 ) ≈ 24.5 m/s

• Angle of projection speed with horizontal = tan –1 (uy /ux) = tan –1 (19.6 /14.8) = 530

• Hence, projection speed of the ball as seen from the road = 24.5 m/s at 530 with horizontal. (Ans)

x

y

ux

uy

u

Kinematics

• Q. Six particles situated at the corner of a regular hexagon of side a move at constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.

• SOLUTION: • Please slide show for this problem.• Click continuously to see animation.

• • All the six colored particles meet at the centre after some time. • We can solve this problem by many methods. • But we shall solve this by concentrating on the motion of two particles only.

Kinematics

• METHOD 1: • Let us concentrate on two particles A & B (say) only. • Initial separation between two particles = side of hexagon = a. • Final separation = 0. • Thus, relative displacement between two particles = a. • Particle B has a component of velocity v cos600 along particle AB. • Thus, relative velocity with which A approaches B each other • = v – v cos600 = v/2 • As v is uniform, therefore time taken by these two balls to meet each other• = (relative displacement)/(relative velocity) = a / (v/2) = 2a/v.

• Hence, the time taken by the particles to meet each other = 2a/v. (Ans)

v

v600

A

B

a

v cos600

Kinematics

P. K. BHARTI, I.I.T. KHARAGPUR

Kinematics

© 2007-10 P K Bharti, IIT Kharagpur