Olympiad Test: Motion- 2 - Class 9 MCQ

Applying Pythagoras

theorem,

OQ = √(OP²+PQ ²)

=√(36+64)=10m

Since,OQ is the actual displacement.

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.

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

• The man first travels 2 m East.
• Next, he moves 6 m South.
• Finally, he goes 6 m East again.
• Total distance East = 2 m + 6 m = 8 m.
• Total distance South = 6 m.
• To find the resultant displacement, use the Pythagorean theorem:
• Resultant displacement = √(8^2 + 6^2) = √(64 + 36) = √100 = 10 m.
• Thus, the correct answer is 10 m.

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.

As the direction keeps on changing continuously in a circular motion, the speed remains constant but velocity keeps on changing all the time.

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 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.

v = 0 u = 20 m/s, t = 5 s; a

=(v-u)/t =(0-20)/5= – 4 m/s²