"For an object moving at a constant speed, we would expect to see a position graph with a
1 answer:
Explanation:
The speed of an object is given by the total distance divided by the total time taken.
The position of an object shows its location. If we draw a position- time graph for an object moving at a constant speed, it will look like the attached figure. It means that the object is moving at a steady rate. It is a straight line passing through origin.
You might be interested in
(a) The launching velocity of the beetle is 6.4 m/s
(b) The time taken to achieve the speed for launch is 1.63 ms
(c) The beetle reaches a height of 2.1 m.
(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.
Use the equation of motion,
Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.
The launching speed of the beetle is <u>6.4 m/s</u>.
(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.
The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.
(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.
Use the equation of motion,
Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.
The beetle can jump to a height of <u>2.1 m</u>
a) 0.94 m
The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:
where:
is the force applied by the snow
d is the displacement of the man in the snow, so it is the depth of the snow that stopped him
m = 68 kg is the man's mass
v = 0 is the final speed of the man
u = 55 m/s is the initial speed of the man (when it touches the ground)
and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)
Solving the equation for d, we find:
b) -3740 kg m/s
The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:
where
m = 68 kg is the mass of the man
is the change in velocity of the man
Substituting,