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

Acceleration = change in velocity/Time taken

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.

u = 0, v = 72 km/h = 72 × (5/18)= 20 ms^–1,

t = 5 minutes=300s

a =(v-u)/t =20/300=1/15 m/s²

=>v²=u²+2as=3000ssms=3kms.

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.

Distance = (½)× base × height=200m.