Olympiad Test: Motion- 1 - Class 9 MCQ

Distance travelled = (½) × Area of rectangle OQPR

=(½) × OR × OQ

OR = speed,

OQ = time at

OP → velocity curve.

v=72km/h=72×(5/18)=20m/s

u=36km/h=36×(5/18)=10m/s

t=10 sec.

A=(v-u)/t=(20-10)/10=1 m/s²

The slope of a velocity-time graph represents acceleration because it quantifies the change in velocity over time. It is calculated as:

• Δv - Change in velocity
• Δt - Change in time

Thus, acceleration can be expressed as:

Acceleration = Δv / Δt

When a body comes to rest, v = 0, here u = 30 m/s,
a = –1.5 m/s²
v = u + at
t=(v-u)/a =(0-30)/-1.5 =20 seconds

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.

Initial velocity = 0, final velocity = 72 km/h = 72 × (5/18) = 20 m/s.

Time = 5 minutes = 5 × 60 = 300 seconds.

Acceleration: 20 = 0 + a × 300, a = 20/300 = 1/15 m/s².

Distance: s = 0 + (1/2) × (1/15) × (300)² = (1/2) × (1/15) × 90000 = 90000/30 = 3000 m = 3 km.

Alternatively: Average speed = (0 + 20)/2 = 10 m/s, distance = 10 × 300 = 3000 m = 3 km.

Slope of the distance–time graph shows speed. The slope of line B is smallest, thus cyclist B is the slowest and cyclist C is the fastest among all four cyclists.

A horizontal line parallel to the time axis on a velocity-time graph represents a constant velocity. This means the car is moving at a steady speed of 10 m/s throughout the entire 25 seconds. Therefore, its velocity at the end of 25 seconds remains the same as its initial velocity, which is 10 m/s.

The area under the speed time graph is the distance covered

The area given is a triangle with height and base 20, and we could use the formula for the area of the triangle.

Distance = 1/2 x base × height = 1/2 x 20 x 20 = 200m.

The formula for final velocity is: v = u + at

Where:

• v = final velocity
• u = initial velocity (15 m/s)
• a = acceleration (2.5 m/s² )
• t = time (2 seconds)
Solution Substituting the values into the formula:
v = 15 m/s + (2.5 m/s²× 2 s)
v = 15 m/s + 5 m/s
v = 20 m/s