# Uniformly Accelerated motion Questions

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```					Uniformly Accelerated Motion Questions

Solutions

Q1)

L - R run        V = s/t = + 604 / 2.19 = + 276 m/s
R - L run        V = s/t = - 604 / 2.22 = - 272 m/s

The signs of the displacements and the velocities are the lesson to be learnt.

Q2)

a =    v    =     (+13) - (+28) = - 5.0 m/s2
t             12 – 9

to = 9 s   vo = +28 m/s                           to =12 s   vo = +13 m/s

Acc. = - 5 m/s2

Q3) Since the spacecraft is slowing down at this stage of motion the
acceleration must be in the opposite direction to the velocity, so a = - 10 m/s2.

s                   a               v                 u                  t
+215,000           - 10 m/s2           ?             + 3250 m/s

To find v we use v2 = u2 + 2 as
v  (3250) 2  210(215000)
v = + 2.5 * 103 m/s or -2.5 * 103 m/s

Both these answers correspond to the same displacement s = 215,000 m.
The + answer corresponds to the point where the craft slows down but is still
travelling in the initial direction. The - answer corresponds to the point where
the retrorocket has caused the ship to stop and then accelerated it to 2.5 * 103
m/s in the opposite direction. It is still at a displacement of 215,000 m from
the initial point but this is the second time it has reached this point.

It takes t = 75 s for the spacecraft to reach the first point in the path and t =
575 to reach the same point travelling in the opposite direction.
Q4) To catch the bus the bull must not only run 11m but also the distance
travelled by the moving bus.

sbus + 11 m = sbull

Assuming we begin at t = 0, the bull is 11m from the bus, his displacement is
sbull = (5) * t

sbus + 11 m = (5) * t

The result reveals that we need a value for sbus to calculate t.

BUS DATA
s                  a                 v                 u            t
?              + 1 m/s2                                0         as bull

sbus = uo t + 1/2 a t2   = 1/2 * ( 1 ) * t2

Substituting, we get         [1/2 * ( 1 ) * t2 ] + 11 m = (5) * t

Solving       (0.5)t2 - (5)t + 11 = 0

we use the quadratic solution to get t = 3.3 s or 6.7 s
[you may need to sing your song at this point]

So there are two times that the bull can catch the bus, the first time is when
he first catches up with it, whilst he is travelling faster than the bus. The
second time is when the bus pulls away at a greater velocity than the bull..
Q5)
a)

s                  a                  v                  u         t
?             - 9.80 m/s2             0               + 30 m/s

So we can use v2 = u2 + 2as to find s.

s = v2 - u2        =    0 - (+30)2       =   45.9 m
2a               2 (9.80)

b)

Method 1
During the time the ball travels upward, gravity causes the speed to decrease
to zero. On the way down, gravity causes the ball to regain this lost speed.
The time for the ball to go up is the same for the time for the ball to come
down. v = u + at can be used to calculate the time taken for the upward
journey.

t =     v-u =         - 30         = 3.06 s
a                        -9.80

Thus t = 2 * 3.06 = 6.12 s

Method 2
It is possible to determine the total time by considering the displacement for
the whole trip to be s =0 m so,

s                  a                  v                  u        t
0             - 9.80 m/s2                             + 30 m/s    ?

Using s = ut + 1/2 at2 we can get       s = (u + 1/2 at)t
0 = (30 + 1/2 *(9.81*t))t
and solving the quadratic we get t = 0 s or 6.12 s.
Q6)
The maximum height that the ball achieves is a characteristic of the vertical
part of the motion. We can ignore the horizontal motion for the moment.

Vertical component of initial velocity ( u y ) = u sin     = 22 ( sin 40o )
= + 14 m/s

The maximum height H can be determined by noting that the y component of
the velocity uy decreases as the ball moves upwards. At the peak of the
motion uy = 0. Using v2 = u2 + 2as we can find s in the y direction.

s                a                  v               u                 t
H=?          - 9.80 m/s2             0            + 14 m/s

s = H =      v2 - u2   =   -( + 14)2         =   + 10 m
2a           2 ( - 9.80)

This is the same height that the ball would have reached had it been thrown
vertically upwards with a velocity of + 14 m/s.

+y

+x

H = maximum
height
Q7) Given the initial velocity, it is the acceleration due to gravity that
determines how long the ball spends in the air. We shall deal with the vertical
motion only. The ball starts at ground level and returns to grounds level and
so the displacement in the y direction is s = 0 m. The initial velocity is the
same as the previous question uy = + 14 m/s.

s              a                v                 u               t
0         - 9.80 m/s2                          + 14 m/s           ?

Using              s = ut + 1/2 a t2 = ( u + 1/2 at) t
0 = [14 + (1/2 * ( -9.80)*t) ] t

Solving with the quadratic formula we get           t = 0 s or 2.9 s

t = 2.9 s

An alternative solution involves calculating the time taken for the ball to reach
maximum height and multiplying by two.

Q8) This time we are to consider the horizontal component of the initial
velocity. We know the time of the motion and so the range s of the ball can be
calculated.

Horizontal component of initial velocity (uv )
= u cos = 22 ( cos 40o )
= + 17 m/s

Range of ball               s   = uv * t        =   17 * 2.9

s = 49 m

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