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Q.1. A body moves so that it follows the following relation dt/dv = –v^{2} + 2v – 1 where v is speed in m/s and t is time in second. If at t = 0, v = 0 then
(a) terminal velocity is 1 m/s
(b) the magnitude of initial acceleration is 1 m/s^{2}
(c) instantaneous speed is v = 1/1 + t
(d) the speed is 1.5 m/s when acceleration is one fourth of its initial value
Correct Answer is Option (a, b and d)
dv/dt = – v^{2} + 2v – 1
Terminal velocity is attained when a = 0
⇒ dv/dt = 0
Q.2. A particle moves in a straight line with an a–t curve shown in figure. The initial displacement and velocity are zero.
(a) The time at which the particle comes to rest again is 10 sec.
(b) Velocity of particle at t = 2 sec is 8 m/s.
(c) Distance travelled by particle before coming to rest is 29.36 m.
(d) Displacement of the particle before coming to rest is 29.36 m.
Correct Answer is Option (a, b c and d)
Area under at curve:
Δv = Area 1 = 2 × 4 = 8
v  u = 8
v = u + 8 = 0 + 8 = 8 m/s
v'  v = Area 2 = (1/2 x 8x2)= – 8 m/s
v' = v – 8 = 8 – 8 = 0
Final velocity is zero at t = 10 sec
Displacement: Can be directly calculated from a–t cure without using v–t curve.
ΔS = u_{0}t_{0 }+ (area under at curve) (t_{0}  t_{c})
Where ΔS = displacement
u_{0} = initial velocity
t_{0 }= total timet_{c} = Abscissa of centroid of corresponding area
Centroid of area 1: C_{1 }= (1,2)
Centroid of area 2 : C_{2} =
= 8 × 3.666 = 8 × 3.67
ΔS = 29.36 m
Q.3. A train travels between two of its station stops with the acceleration schedule shown. Δt is the time interval during which train breaks to stop.
(a) Δt = 10 sec
(b) Distance between two stations is 416 m
(c) Δt = 15 sec
(d) Distance between two stations is 510 m
Correct Answer is Option (a and b)
Area under graph gives change in velocity
Train starts from rest and comes to rest
Therefore, Positive area = negative area
8 × 1 + 6 × 2 = 2 × Δt
8 + 12 = 2Δt
Δt = 20 = 10 sec
Displacement & distance are equal in this case displacement can be directly calculated from at curve without using vt curve as follows
ΔS = ut_{0} + (area under at curve) (t_{0}  t_{c})
Where u → initial velocity
t_{0} → total time
t_{c} → abscissa of centroid of corresponding area
Δs = 8 [34  4] + 12 [34  11] + (20) [34  29]
= 8 × 30 + 12 × 23  20 × 5
= 240 + 276 – 100
= 516 – 100
Δs = 416 m
Q.4. A particle moves with an initial velocity v_{0} and retardation αv, where v is its velocity at any time t
(a) The particle will cover a total distance v_{0}/α
(b) The particle will come to rest after a time 1 /α
(c) The particle will continue to move for a very long time
(d) The velocity of the particle will become v_{0}/2 after a time 1/α
Correct Answer is Option (a and c)
Q.5. In the figure is shown the position of a particle moving on the xaxis as a function of time. Then 
(a) the particle has come to rest for 6 times
(b) the maximum speed is at t = 6 s
(c) the velocity remain positive for t = 0 to t = 6s
(d) the average velocity for the total period shown is negative.
Correct Answer is Option (a and d)
In x  t graph at maxima and minima. dt/dx = 0, hence particle comes to rest for 6 times and average velocity for total period is negative.
Q.6. A projectile is projected with a speed of 40m/s at an angle θ with horizontal such that
tan θ = 3/4. After 2 sec, the projectile is moving with speed v at an angle α with horizontal then, (g = 10 m/s^{2})
(a) tan α = 1/8
(b) cot α = 1/8
(c) v = 32.25 m/s
(d) 32 m/s
Correct Answer is Option (a and c)
and v cos α = u cos θ
∴ v x 8/√65 = 40 x 4/5 = 32.25 m/s
Q.7. Path of three projectiles are shown. If T_{1}, T_{2} and T_{3} are time of flights and ignoring air resistances –
(a) T_{1} > T_{3}
(b) T_{1} < T_{3}
(c)
(d) T_{1} = T_{2} = T_{3}
Correct Answer is Option (c and d)
As maximum heights are same therefore their time of flights will be same.
Q.8. Two particles are projected from the same point with the same speed, at different angles θ_{1} and θ_{2} to the horizontal. Their times of flight are t_{1} and t_{2 }and they have the same horizontal range. Then
(a) t_{1}/t_{2} = tan θ_{1}
(b) t_{1}/t_{2 }= tan θ_{2}
(c)
(d) θ_{1} + θ_{2} = 90º
Correct Answer is Option (A, c and d)
For same range, θ_{1} and θ_{2}.
θ_{2 }= (90  θ_{1}) ⇒ θ_{1} + θ_{2} = 90º
Q.9. A body is thrown from a point with speed 50 m/s and an angle 37º with horizontal in time t it has moved a horizontal distance of 80 m, then its distance from point of projection is d, then – (g = 10 m/s^{2})
(a) d = 40 m
(b) 40√5 m
(c) t = 2 sec
(d) t = 2.67 sec
Correct Answer is Option (a and b)
Q.10. A particle is projected with a speed v and an angle θ with horizontal. Choose the correct statements 
(a) Speed of the particle is zero at highest point.
(b) Speed of the particle is minimum at highest point.
(c) Acceleration of particle throughout the motion is constant.
(d) Acceleration of particle varies with time.
Correct Answer is Option (b and c)
At highest point, speed is not zero and it is equal to u cos θ.
Acceleration is constant.
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