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The quantity which is measured by the area occupied under the speedtime graph is
Distance travelled = (½) × Area of rectangle OQPR
=(½) × OR × OQ
OR = speed,
OQ = time at
OP → velocity curve.
A bus increases its speed from 36 km/h to 72 km/h in 10 seconds. Its acceleration is
v=72km/h=72×(5/18)=20m/s
u=36km/h=36×(5/18)=10m/s
t=10 sec.
A=(vu)/t=(2010)/10=1 m/s^{2}
Acceleration = change in velocity/Time taken
An object moving with a velocity of 30 m/s decelerates at the rate of 1.5 m/s2. Find the time taken by the object to come to rest
When a body comes to rest, v = 0, here u = 30 m/s,
a = –1.5 m/s^{2}
v = u + at
t=(vu)/a =(030)/1.5 =20 seconds
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.
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find the distance travelled by the train for attaining this velocity
u = 0, v = 72 km/h = 72 × (5/18)= 20 ms^{–1},
t = 5 minutes=300s
a =(vu)/t =20/300=1/15 m/s^{2}
=>v^{2}=u^{2}+2as=3000ssms=3kms.
Four cyclists A, B, C, and D are cycling on a levelled straight road. Their distance–time graphs are shown in the given figure. Which of the following is correct regarding the motion of these cyclists?
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 car of mass 1000 kg is moving with a velocity of 10 ms–1. If the velocity–time graph for this car is a horizontal line parallel to the time–axis, then the velocity of car at the end of 25 s will be
A horizontal line parallel to the time–axis shows a uniform velocity throughout the motion of the object.
What is the distance covered by a particle during the time interval of 20 seconds, for which the speed–time graph is shown in the adjacent figure
Distance = (½)× base × height=200m.
A bus moving along a straight line at 15 m/s undergoes an acceleration 2.5 m/s2 . After 2 seconds,its speed will be,
v = u + at
=20 m/s.
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