DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following.
Statement 1 :In a free fall, weight of a body becomes effectively zero.
Statement 2 :Acceleration due to gravity acting on a bodyhaving freefall is zero.
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Solution
(d) Statement -1 is true, Statement-2 is false
DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following.
Statement 1 :A positive acceleration can be associated with a ‘slowing down’ of the body.
Statement 2 :The origin and the positive direction of an axis area matter of choice.
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Solution
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following.
Statement 1 : Velocity-time graph for an object in uniform motion along a straight path is a straight line parallel to the time axis.
Statement 2 : In uniform motion of an object velocity in creases as the square of time elapsed.
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Solution
(d) Statement -1 is true, Statement-2 is false
In uniform motion the object moves with uniform velocity,the magnitude of its velocity at different instane i.e., at t = 0, t =1, sec, t= 2 sec .....will always be constant.Thus velocity-time graph for an object in uniform motion along a straight path is a straight line parallel to time axis.
The displacement of particle is given by \(x=a_{0}+\frac{a_{1}t}{2}-\frac{a_{2}t^{2}}{3}\). What is its acceleration?
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Solution
A particle moves along a straight line OX. At a time t(in second) the distance x (in metre) of the particle from O is given by x = 40 + 12t –t3. How long would the particle travel before coming to rest?
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Solution
A metro train starts from rest and in 5 s achieves 108 km/h.After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.
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Solution
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
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Solution
Let O be the origin, then
passenger in the train at P observes the car at Q along the direction PQ; i.e.west north direction.
Velocity time curve for a body projected vertically upwards is
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Solution
Velocity time curve will be a straight line as shown:
At the highest point v = 0.
A car travelling at a speed of 30 km h–1 is brought to a halt in 8 m by applying brakes. If the same car is travelling at 60 kmh–1, it can be brought to a halt with the same braking power in
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Solution
A train A which is 120 m long is running with velocity 20 m/s while train B which is 130 m long is running in opposite direction with velocity 30 m/s. What is the time taken by train B to cross the train A ?
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Solution
Here, length of train A, LA= 120 m
length of train B, LB= 130 m
velocity of train A, vA= 20 m/s
velocity of train B, vB= 30 m/s
Train B is running in opposite direction to train B,
∴ velocity of train B relative to train A,
vBA= vB + vA
= (30 + 20) m/s
= 50 m/s
Total distance to be covered by train B
= LA + LB= (120 + 130) m= 250 m
Hence, time required by train B to cross train A
t= \(\frac{250}{50}\) sec = 5 sec