GSC101 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || GENERAL SCIENCE || BY VuTech
Question-1
Explain the motion of an object in the following Situations with suitable examples from daily life,
a)
Speed is constant but velocity is
varying
b)
Acceleration of a body is constant but velocity is zero
c)
Speed is constant but acceleration is not zero
d) Force is applied but acceleration is zero
Solution:
a) Speed is constant but velocity is varying
In this situation, the object is moving at a constant speed, but its
direction is changing. This can be seen in the motion of a car turning a
corner. The car's speed may be constant, but its velocity is changing because
its direction is changing.
b) Acceleration of a body is constant but velocity is zero
In this situation, the object is not moving, but it is accelerating. This
can be seen in the motion of a car that is stopped at a red light. The car's
velocity is zero, but it is accelerating because the force of the engine is
still acting on it.
c) Speed is constant but acceleration is not zero
In this situation, the object is moving at a constant speed, but it is
accelerating. This can be seen in the motion of a car that is moving in a
straight line at a constant speed. The car's speed is constant, but it is
accelerating because the force of the engine is still acting on it.
d) Force is applied but acceleration is zero
In this situation, a force is applied to an object, but the object does not accelerate. This can be seen in the motion of a car that is stuck in mud. The force of the engine is still acting on the car, but the car does not accelerate because the mud is providing a force that is equal to the force of the engine.
KINDLY, DON’T COPY PASTE
SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES
Question-2
The
required take off speed of a commercial aeroplane is 120 knots. If this
commercial aeroplane take 30-seconds to covers the airport runway path,
calculate the
a) Acceleration required during take off
b) Runway length of airport
[Hint: knot is nautical mile per hour or equal
to half a meter per seconds]
Solution:
a) Acceleration required during take off
The
acceleration required during take off can be calculated using the following
formula:
a = tvf − vi
where:
- a is the acceleration
- vf is the final velocity
- vi is the initial velocity
- t is the time
In this case,
the final velocity is 120 knots, the initial velocity is 0 knots, and the time
is 30 seconds. Substituting these values into the formula, we get:
a = 30 seconds120 knots−0 knots
a = 4 knots/second
Converting
knots to meters per second, we get:
a = 4 knots/second × 1 knot/second 0.51444 meters/second
a = 2.05776 meters/second2
Therefore,
the acceleration required during take off is 2.05776 meters per second squared.
b) Runway length of airport
The runway
length of the airport can be calculated using the following formula:
d = 2avf2 − vi2
where:
- d is the distance traveled
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration
In this case,
the final velocity is 120 knots, the initial velocity is 0 knots, and the
acceleration is 2.05776 meters per second squared. Substituting these values
into the formula, we get:
d = 2 × 2.05776 meters/second2(120 knots)2−(0 knots)2
d = 3600 meters
Therefore, the runway length of the airport is 3600 meters.