Q#12
A particle starts from point A and travels along the solid curve shown in figure (3-E7). Find approximately the position B of the particle such that the average velocity between the positions A and B have the same direction as the instantaneous velocity at B.
Answer:
Instantaneous velocity at B will be along the tangent at point B, while the direction of average velocity will be along the direction of displacement AB.
So, both to have the same direction we draw a line from point A to curve such that it is tangent to the curve. The point where it is tangent to the curve is the required point B. It has been shown in the following figure:
Q#13
An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance traveled during the period of acceleration.
Answer:
Initial velocity u = 4.0 m/s²
acceleration a = 1.2 m/s².
time t = 5.0 s
from the equation of accelerated motion on a straight line
x = ut + ½at²
x = 4 (5) + ½ (1.2)(5)² = 35 m
Q#14
A person traveling at 43.2 km/h applies the break giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?
Answer:
We should convert the given data in matching units.
u = 43.2 km/h = 43.2(1000/3600) m/s = 12 m/s
a = -6.0 m/s², Final velocity v = 0 m/s
from the equation of accelerated motion on a straight line
v² = u² + 2ax
0² = (12)² - 2(6)x
x = (12)²/12 = 12 m
Q#15
A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.
Answer:
We solve it in two parts 1: accelerated motion and 2 decelerated motion.
For 1: accelerated motion
u = 0, a = 2.0 m/s². t = ½(60) = 30 seconds, Now final velocity is given by v = u + at = 0 + 2(30) = 60 m/s
Since after this velocity brakes are applied, so it is the
(b) maximum speed attained = 60 m/s.
Distance traveled x = ut + ½at² = 0 + ½(2)(30)² = 0.9 km
Let the distance traveled at half the maximum speed (30 m/s) be y, Now from the equation
v²= u² + 2ax,
30² = 0² + 2(2)y
y = 900/4 = 225 m (1)
For 2: decelerated motion
u = 60 m/s, v = 0 m/s, t = 60 s From the equation v = u + at
a = (0 – 60)/60 = -1 m/s²
Distance traveled = ut + ½ at²
= 60(60) + ½(-1)(60)² = 3600 – 1800 = 1.8 km
Hence answer for (a) The total distance moved = 0.9 km + 1.8 km
= 2.7 km
Let the distance moved at half the speed in this part be z
From v² = u² + 2ax
30² = 60² + 2(-1)z
2z = 60² - 30² = 2700 m
z = 1350 m = 1.35 km
(c) There will be two positions at half the maximum speed (From the point of start),
From (1) = 225 m and from (2) 1.35 km + 0.9 km = 2.25 km
Q#16
A bullet traveling with a velocity of 16 m/s penetrates a tree trunk and comes to rest 0.4 m. Find the time taken during the retardation.
Answer:
Initial velocity u = 16 m/s, distance moved = 0.4 m, final velocity =0 m/s From the equation of motion v² = u² + 2ax
0 = 16² + 2(a)0.4
a = -16²/0.8 = -320 m/s (-ve sign is for retardation)
Time taken during the retardation is given by the equation v = u + at
0 = 16 – 320t
t = 0.05 s
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