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A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/h and 40 km/h respectively. The velocity of the car midway between P and Q is [1988]
Let PQ = x , then
Also, velocity at mid point is given by
This gives
A train of 150 metre long is going towards north direction at a speed of 10 m/s . A parrot flies at the speed of 5 m/s towards south direction parallel to the railway track. The time taken by the parrot to cross the train is [1988, 1992]
Relative velocity of parrot w.r.t the train = 10 – (–5) = 15 ms^{–1}.
Time taken by parrot to cross the train
What will be the ratio of the distances moved by a freely falling body from rest in 4th and 5th seconds of journey? [1989]
A car covers the first half of the distance between two places at 40 km/h and other half at 60 km/h.The average speed of the car is [1990]
Total distance = s;
Total time taken =
∴ Average speed =
A bus travelling the first one third distance at a speed of 10 km/h, the next one third at 20 km/h and the last onethird at 60 km/h. The average speed of the bus is [1991]
Let total distance be x
It covered
x/3 at 10
x/3 at 20
x/3 at 60
Times taken respectively are { d / t }
x / 30
x / 60
x / 180 respectively
total time = x/30 + x /60 + x/180
= 10x / 180
= x / 18
Average speed of bus (avg s )
= total distance / total time
avg s = x / {x/18}
avg s = 18 km / h
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. Find the value of v
or v = 60 km/h
⇒ v = 60 km/h
A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s^{2}) [1991]
Let the body fall through the height of tower in t seconds. From,
we have, total distance travelled in last 2 seconds of fall is
D = D_{t}+D_{(t 1)}
or, 40 = 20 (t – 1) or t = 2 + 1 = 3s
Distance travelled in t seconds is
Which of the following curve does not represent motion in one dimension? [1992]
In one dimensional motion, the body can have at a time one velocity but not two values of velocities.
A body starts from rest, what is the ratio of the distance travelled by the body during the 4th and 3rd seconds ? [1993]
u = 0 m/s
S = u + (1/2)a(2n  1)
S₄ = 0 + 1/2(2×4  1)
= 7a/2
S₃ = 0 + (1/2) a (2×3  1)
= 5a/2
Ratio, S₄/S₃ = 7a/2 ÷ 5a/2
=7/5
A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is [1994]
The displacement time graph of a moving particle is shown below
The instantaneous velocity of the particle is negative at the point [1994]
At E, the slope of the curve is negative.
A particle moves along a straight line such that its displacement at any time t is given by s = (t^{3} – 6t^{2 }+ 3t + 4) metres The velocity when the acceleration is zero is
Velocity,
Acceleration,
we have, 0 = 6 t – 12 or t = 2s.
Hence, at t = 2 s the velocity will be
A body is thrown vertically upward from the ground. It reaches a maximum height of 20 m in 5 sec. After what time, it will reach the ground from its maximum height position? [1995]
h_{max} = 20 m and t = 5 sec. Time taken by the body to reach the ground from some height is the same as taken to reach that height.
Therefore, time to reach the ground from its maximum height is 5 sec.
A stone released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is (g = 10m / s^{2} ) [1995]
Initial velocity (u) = 0; Time (t) = 4 sec and gravitational acceleration (g) = 10 m/s^{2}.
Height of tower
= 80 m.
Three different objects of masses m1, m2 and m3 are allowed to fall from rest and from the same point O along three different frictionless paths.The speeds of the three objects on reaching the ground will be in the ratio of [1995]
The speed of an object, falling freely due to gravity, depends only on its height and not on its mass. Since the paths are frictionless and all the objects fall through the same height, therefore, their speeds on reaching the ground will be in the ratio of 1 : 1 : 1.
The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant ? (Take g = 10 m/s^{2})[1995]
Height of tap = 5m and (g) = 10 m/sec^{2}.
For the first drop, 5 =
= 5t^{2 }or t^{2} = 1 or t = 1 sec.
It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.
Distance covered by the second drop in 0.5 sec
= 1.25 m.
Therefore, distance of the second drop above the ground = 5 – 1.25 = 3.75 m.
If a ball is thrown vertically upwards with a velocity of 40 m/s, then velocity of the ball after two seconds will be (g = 10 m/s^{2}) [1996]
Initial velocity (u) = 40 m/s
Acceleration a = –g m/s^{2} = –10 m/s^{2}
Time = 2 seconds
By Ist equation of motion, v = u + at
v = 40 – 10 (2) = 20 m/s
The displacement of a particle varies with time (t) as: s = at^{2} – bt^{3}. The acceleration of particle is zero, at time T equal to
s = at^{2 }– bt^{3}
If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of [1997]
Initial velocity of car (u) = 0
Final velocity of car (v) = 144 km/hr = 40 m/s
Time taken = 20 s
We know that, v = u + at 40 = a × 20 ⇒ a = 2 m/s^{2}
Also, v^{2} – u^{2} = 2as
A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? [1998]
v^{2} – u^{2} = 2as
where u_{1} = 40 km/h
For same retarding force _{s ∝ u}^{2}
∴ s^{2} = 4s_{1 } = 8mIf
F is retarding force and s the stopping distance, then
For same retarding force, s α v^{2}
∴ s_{2} = 4s_{1} = 4 x 2= 8m
The displacement of a particle is represented by the following equation : s = 3t^{3} + 7t^{2} + 5t+8 where s is in metre and t in second. The acceleration of the particle at t = 1s is [2000]
Displacement s = 3t^{3} + 7t^{2} + 5t +8;
Velocity =
Acceleration =
Acceleration at (t = 1s) = 18×1+14 = 18 + 14 = 32 m/s^{2}
A stone is thrown vertically upwards with kinetic energy K. The kinetic energy at the highest point is [2001]
At heigh est point, velocity is zero.
Hence, Kinetic energy is zero.
If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is [2003]
Let body takes T sec to reach maximum height.
Then v = u – gT
v = 0, at highest point.
...................(1]
Velocity attained by body in (T – t) sec
v = u – g (T – t)
or v = gt ...................(2)
∴ Distance travelled in last t sec of its ascent
A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? [Given g = 9.8 m/s^{2}] [2003]
Let the required speed of throw be u ms^{–1}.
Then time taken to reach maximum height,
For two balls to remain in air at any time, t must be greater than 2.
The displacement x of a particle varies with time t as x = ae^{αt} + be^{βt}, where a, b, α and β are positive constants. The velocity of the particle will [2 00 5]
Given x = ae^{–αt} + be^{βt}
Velocity, v =
i.e., go on increasing with time.
A ball is thrown ver tically upward. It has a speed of 10 m/sec when it has reached one half of its maximum height. How high does th e ball rise? [2001, 2005] Take g = 10 m/s^{2}.
For part AB
From 3rd equation of motion v^{2} = u^{2} – 2gH
0 = u^{2} – 2g(H/2) = u^{2} – gH
A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t – t^{3}. How long would the particle travel before coming to rest? [2006]
At t = 0, the particle is at, let's say x distance, from O;
then putting t = 0 in the given displacementtime equation we get;
► x = 40 + 12(0) − (0)^{3} = 40m
Particle comes to rest that means the velocity of the particle becomes zero after travelling certain displacement; let's say the time be t.
then after differentiating the given displacement−time equation wrt. time we get the velocity−time equation
► v = 12 − 3t^{2}
at time t = t (the time when the particle comes to rest ): v = 0;
► 12 − 3t^{2 }= 0;
► t = 2s
Then, at t = 2s we are at, let's say x′ distance from O;
put this value of t (=2) in given displacementtime equation, we get;
► x′ = 40 + 12 (2) − (2)^{3};
= 56m
Further;
We have seen that the particle started his journey when it is at 40m from point O.
And came to rest at 56m from point O.
then the particle travelled a distance of:
► 56 − 40 = 16m.
Two bodies, A (of mass 1 kg) and B (of mass 3 kg), are dropped from heights of 16m and 25m, respectively. The ratio of the time taken by them to reach the ground is [2006]
Let t_{1 }& t_{2} be the time taken by A and B respectively to reach the ground then from the formula,
For first body,
For second body,
The position x of a particle with respect to time t along xaxis is given by x = 9t^{2} – t^{3} where x is in metres and t in second. What will be the position of this particle when it achieves maximum speed along the +ve x direction? [2007]
Speed
For maximum speed
⇒ x_{max }= 81  27= 54
A car moves from X to Y with a uniform speed v_{u} and returns to Y with a uniform speed v_{d}. The average speed for this round trip is [2007]
Average speed
Let s be the distance from X to Y.
∴ Average speed
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