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A bus starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Given- u=0 (start from rest)
a= 0.1 m/s2, t= 2 min= 2x60 sec =120 sec
To find- a) v=? & b) s=?
Solution- a) v = u +at= 0+ 0.1x 120 = 12m/s
b) s= ut + 1/2 a (t) ² = 0 + 1/2 x0. 1x (120) ² = 720 m
Concept- Always covert all quantities in to same unit ( prefer SI units)
A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.
Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
acceleration , a = -0.5 m/s²
Use formula,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50 sec
Again, use formula,
S = ut + 1/2at²
Where S is distance travelled before stop
S = 25 × 50 - 1/2 × 0.5 × 50²
= 1250 - 1/2 × 0.5 × 2500
= 1250 - 625
= 625 m
Hence, distance travelled = 625m and time taken 50 sec
A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?
In 1 sec Velocity of body increases by 2cm/sec. So in 3sec it's velocity would be 2cm/sec 3= 6cm/s
A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?
if u=0 then S= ut+1/2 at^2 s=1/2*4*100= 200m
What can you say about the motion of an object whose distance−time graph is a straight line parallel to the time axis?
When an object is at rest, its distance−time graph is a straight line parallel to the time axis.
A straight line parallel to the x-axis in a distance−-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.
What can you say about the motion of an object if its speed−time graph is a straight line parallel to the time axis?
Object is moving uniformly.
A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.
What is the quantity which is measured by the area occupied below the velocity−time graph?
Distance
The graph shows the velocity−time graph of a uniformly moving body.
Let the velocity of the body at time (t) be v.
Area of the shaded region = length × breath
Where,
Length = t
Breath = v
Area = vt = velocity × time …(i)
We know,
Velocity = DisplacementTIme Velocity = DisplacementTIme
∴ Displacement = Velocity × Time…(ii)
From equations (i) and (ii),
Area = Displacement
Hence, the area occupied below the velocity−time graph measures the displacement covered by the body.
What is the nature of the distance−time graphs for uniform and non-uniform motion of an object?
The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure).
The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure).
A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m s−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
1.25m
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