Q.1. A particle is moving along the xaxis, with its coordinate as a function of time given by x(t) =10+ 8t  3t^{2}. Another particle is moving along the yaxis with its coordinate as a function of time given by y(t) = 5  8t^{3}. At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as √v. Then v (in m/s) is ______. [2020]
Ans. 580
Solution: Given that
x(t) =10+ 8t  3t^{2}, y(t) = 5  8t^{3}
So,
At t = 1 s, we have
Relative velocity of second particle with respect to first particle is
Q.2. A particle moves such that it's position vector is where ω is a constant and t is time. Then which of the following statements is true for the velocity and acceleration of the particle.
(1) is perpendicular to is directed away from the origin.
(2) both are perpendicular to
(3) both are parallel to
(4) is perpendicular to is directed towards the origin.
Ans: (4)
Solution: Given that
So,
and
Since, is perpendicular to
And had opposite direction to or antiparallel to that is, directed towards the origin.
Q.3. A ball is dropped from the top of a 100 m high tower on a planet. In the last (1/2)s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in m/s^{2}) near the surface on that planet is _______.
Ans. 8
Solution: Given that
u = 0 m/s, H = 100 m, d = 19 m, t = 1/2 s
Let time taken to tavel (H – h) distance be t.
So,
..(1)
Now, time taken to hit the ground is
..(2)
From Eqs. (1) and (2), we get
⇒ a = 8 m/s^{2}
Q.4. A particle starts from the origin at t = 0 with an initial velocity of and moves in the xy plane with a constant acceleration The xcoordinate of the particle at the instant when its ycoordinate is 32 m is D m. The value of D is
(1) 32
(2) 50
(3) 60
(4) 40
Ans. (3)
Solution: Given that
From the equation of motion in two dimensions, we have
⇒ t^{2} = 16
⇒ t = 4 s
So,
Q.5. The distance x covered by a particle in one dimensional motion varies with time t as x^{2} = at^{2} + 2bt + c. If the acceleration of the particle depends on x as x^{−n}, where n is an integer, the value of n is ______.
Ans. 3
Solution: x^{2} = at^{2} + 2bt + c
.. (1)
Velocity is given by
⇒ vx = at + b ..(2)
Acceleration is given by
Differentiate Eq. (2) with respect to t, we get
So,
Therefore, n = 3
Q 6. In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed v more than that of car B. Both the cars start from rest and travel with constant acceleration a_{1} and a_{2}, respectively. Then v is equal to [JEE Main 2019]
(1)
(2)
(3)
(4)
Ans: (3)
Solution: Let time taken by car A is t'
v_{B} = a_{2}(t' +t)
Put this value in Eq. (1), we get
Q 7. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of the. The ratio of times taken by the passenger train to completely cross the freight train when (i) they are moving in same direction and (ii) in the opposite directions is [JEE Main 2019]
(1) 11/5
(2) 5/2
(3) 3/2
(4) 25/11
Ans: (1)
Solution:
Let passenger train and freight train be A and B and v_{A} is the velocity of passenger train and v_{B} is the velocity of freight train B.
Relative velocity of A with respect to B is (same direction)
Time taken by the train A to cross the train B
Train moving in opposite direction
Therefore, required ratio is
Q 8. A particle starts from the origin at time t = 0 and moves along the positive xaxis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5 s? (JEE Main 2019)
(1) 10 m
(2) 6 m
(3) 3 m
(4) 9 m
Ans: (4)
As we know that, area under vt curve gives displacement.
therefore,
Q 9. A particle is moving with a velocity where K is a constant. The general equation for its path is (JEE Main 2019)
(1) y = x^{2} + constant
(2) y^{2} = x + constant
(3) y^{2} = x^{2} + constant
(4) xy = constant
Ans: (3)
Solution: Equation for velocity of the particle is
Thus,
Now,
Integrating both the sides, we get
y^{2} = x^{2} + constant
Q 10. The position coordinates of a particle moving in a 3D coordinate system is given by
x = a cos ωt
y = a sin ωt
and
z = aωt
The speed of the particle is (JEE Main 2019)
(1)
(2) aω
(3)
(4) 2aω
Ans: (1)
Solution:
Velocity coordinates of particle is obtained by differentiating the position coordinate with respect to time, Thus:
Hence, velocity of particle is
Q 11. In the cube of side a, as shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be (JEE Main 2019)
(1)
(2)
(3)
(4)
Ans: (3)
Solution:
Side of a cube is a
Q 12. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is (JEE Main 2019)
(1) 1 : 16
(2) 1 : 2
(3) 1 : 4
(4) 1 : 8
Ans: (1)
Solution:
Area covered = πR^{2}
A ∝ R
Thus, range
Therefore, required ratio is
Q 13. Two vectors have equal magnitudes. The magnitude of
is n times the magnitude of The angle between is (JEE Main 2019)
(1)
(2)
(3)
(4)
Ans: (1)
Solution:
We know that
According to the question
Squaring both the sides, we get
Q 14. A body is projected at t = 0 with a velocity 10 m/s at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1 s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 m/s2, the value of R is (JEE Main 2019)
(1) 2.5 m
(2) 2.8 m
(3) 10.3 m
(4) 5.1 m
Ans: (2)
Solution:
At t = 1:
The radius of curvature of the trajectory is
= 2.77 m = 2.8 m
Q 15. A particle moves from the point at t = 0 with an initial velocity It is acted upon by a constant force which produces a consult acceleration What is the distance of the particle from the origin at time 2s? (JEE Main 2019)
(1) 15m
(2) 20√2m
(3) 5m
(4) 10√2m
Ans: (2)
Solution:
Q 16. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. (JEE Main 2018)
(1)
(2)
(3)
(4)
Ans: B
Solution: In graph ‘2’ initial slope is zero which is not possible, since initial velocity is nonzero in all other three graphs.
Q 17. An automobile, travelling at 40 km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding): (JEE Main 2018)
(1) 75 m
(2) 160 m
(3) 150 m
(4) 100 m
Ans: (2)
Solution:
Q 18. The velocitytime graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively. (JEE Main 2018)
(1) 337.5 m and 25 s
(2) 225.5 m and 10 s
(3) 112.5 m and 15 s
(4) 112.5 m and 22.5 s
Ans: D
Solution:
The distance traveled in 15 seconds by both will be given by area under curve.
Suppose after time t, they will meet, then distance traveled by both of them will be equal.
Q 19. Let The magnitude of a coplanar vector is given by: (JEE Main 2018)
(1)
(2)
(3)
(4)
Ans: C
Solution:
Solving equation (i) and (ii) we get
Q 20. Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity? (JEE Main 2017)
Ans: (3)
Solution:
Q 21. A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s^{2} and the car has acceleration 4 m/s^{2}. The car will catch up with the bus after a time of (JEE Main 2017)
(1)
(2) 15 s
(3)
(4)
Ans: (4)
Solution:
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