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If a body moves 6 m towards South and then turns towards East and moves 8 m, then find the displacement of the body
Applying Pythagoras
theorem,
OQ = √(OP^{2}+PQ ^{2})
=√(36+64)=10m
Since,OQ is the actual displacement.
In the speedtime graph for a moving object shown here, the part which indicates uniform
deceleration of the object is
As we move along the graph, PQ and ST show uniform acceleration. QR shows acceleration at an increasing rate. RS shows deceleration i.e.,speed is decreasing.
An artificial satellite is moving in a circular orbit of radius 32,000 km. If it takes 30 hours to complete one revolution around the earth, then find the velocity of the satellite
R = 32000 km, t = 30 hours
Circumference of the orbit = 2πR
=2*(22/7)*32000
=201142.86km
Velocity=d/t=2πR/t=6704 km/h
A man travels a distance of 2 m towards East, 6 m towards South and finally 6m towards the East. The resultant displacement is
A man starts from O and reaches Q. Shortest distance is OQ.
Apply Pythagoras theorem to
∆ORQ.
(RQ = PM) and (PR = MQ)()
OQ^{2}= OR^{2}+ RQ^{2}
OQ=√{(2+6)^{2}+6^{2}}
OQ=10m
The area under a speed–time graph represents a physical quantity which has the unit
Area of trapezium ={(OB+AC)*OA}/2
= Distance travelled in A seconds, whose unit is m. Initial speed = OB, then accelerating from B to C in time A.
The motion in which a body has a constant speed but not constant velocity is called
As the direction keeps on changing continuously in a circular motion, the speed remains constant but velocity keeps on changing all the time.
What can you say about the motion of the body if its speed–time graph is a straight line parallel to the time axis ?
In the case of the speed–time graph, having a constant line parallel to the time (x) axis, we see that speed is fixed (constant) but time keeps on moving in this graph as shown,Speed = u (throughout).
The figure shows distance–time graphs of two cars A and B running at different speeds. Which car is running with a greater speed in comparison to the other car?
The line graph for car B makes a larger angle with time –axis. Its slope is larger than the slope of the line for car A. And the slope of distance–time graph shows speed. Thus car B has greater speed than car A.
A car accelerates from 15 km/h to 60 km/h in 300 seconds. Find the distance travelled by the car during this time
u = 15 km/h, v = 60 km/h, t = 300 sec =1/12 h
a=(vu)/t =(6015)/(1/12)=540 km/h^{2}
Distance travelled by the car is given by
S=ut+(1/2)at^{2}
=15*(1/12) + (1/2)*540*(1/2)
=5 km
A motorcycle is being driven at a speed of 20 m/s when brakes are applied to bring it to rest in five seconds. The deceleration produced in this case will be
v = 0 u = 20 m/s, t = 5 s; a
=(vu)/t =(020)/5= – 4 m/s^{2}
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