Acceleration
Trending Questions
- 8 m/s2
- 12 m/s2
- 15 m/s2
- 4 m/s2
- a0
- a1
- 2a2
- a2
Can the velocity be 0 but acceleration be non zero?
Yes
No
- In space
- More information needed
- 1 m/s2
- 2 m/s2
- 0.5 m/s2
- 1.5 m/s2
This section contains 1 Assertion-Reason type question, which has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
इस खण्ड में 1 कथन-कारण प्रकार का प्रश्न है, जिसमें 4 विकल्प (a), (b), (c) तथा (d) दिये गये हैं, जिनमें से केवल एक सही है।
A : एक वस्तु का तब भी त्वरण हो सकता है, जब इसकी चाल एकसमान हो।
R : If velocity is constant, then there is no change in direction of motion.
R : यदि वेग नियत है, तब गति की दिशा में कोई परिवर्तन नहीं होता है।
- Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) तथा (R) दोनों सही हैं तथा (R), (A) का सही स्पष्टीकरण है - Both (A) and (R) are true but (R) is not the correct explanation of (A)
(A) तथा (R) दोनों सही हैं लेकिन (R), (A) का सही स्पष्टीकरण नहीं है - (A) is true but (R) is false
(A) सही है लेकिन (R) गलत है - (A) is false but (R) is true
(A) गलत है लेकिन (R) सही है
- the terminal speed is 2 m/s.
- the speed varies with the time as v=2(1−e−3t) m/s.
- the speed is 0.1 m/s when the acceleration is half the initial value.
- the magnitude of the initial acceleration is 6 m/s2.
(i) Acceleration at 2 sec.
(ii) Average acceleration in 2 sec.
- (i) 71 m/s2
(ii) 17 m/s2 - (i) 81 m/s2
(ii) 18 m/s2 - (i) 91 m/s2
(ii) 19 m/s2 - (i) 61 m/s2
(ii) 16 m/s2
- 2 αv3
- 2 βv3
- 2 αβv3
- 2 β2v3
- x=αt−13βt3
- x=αt−12βt2
- α=−βt
- Maximum positive displacement from origin =2α3√αβ
- 1√2 m/s2 North-East.
- 12 m/s2 North-West.
- 1√2 m/s2 North-West.
- Zero.
Acceleration of the particle at t=0 is
- e4 m/s2
- 4e4 m/s2
- 4 m/s2
- 0
(Given that at t=0, u=3 m/s, x=0 m)
- (t2+3), (t33+3t22+3t)
- (t2+3t), (t33+3t22)
- (2t2+3t+3), (2t33+3t22+3t)
- (2t2+3t), (2t33+3t22)
- +6√2 m/s
- −6√2 m/s
- 72 m/s
- 0
- 8 m/s2
- 12 m/s2
- 15 m/s2
- 4 m/s2
Column - IColumn - IIi.Velocity of a particle following (p) αequation x=u(t−8)+α(t−2)2at t=0ii.Acceleration of a particle moving according to equation(q)2αx=u(t−8)+α(t−2)2 at t=0iii.Velocity of the particle moving (r) uaccording to equation x=ut+5αt2ln(1+t2)+αt2at t=0iv.Acceleration of the particle moving (s) u−4αaccording to the equation x=ut+5αt2(1+t2)+αt2at t=0
- i→(s);ii→(q);iii→(r);iv→(q)
- i→(s);ii→(q);iii→(s);iv→(p)
- i→(r);ii→(p);iii→(s);iv→(q)
- i→(r);ii→(r);iii→(r);iv→(q)
(Given, at t=0, u=2 m/s, x=2 m)
- 2t33−3t2+2t+2
- t33−3t2+2t
- t33−3t2+2t+2
- t33−3t2
which of the following condition is correct if the particles speed is reducing?
- xvx+yvx<0
- xvx+yvy+yvy>0
- axvx+ayvy<0
- axvx+ayvy>0
The value of dot product of velocity and acceleration of particle at t=2 s is
- 1 m2/s3
- −1 m2/s3
- 2 m2/s3
- −2 m2/s3
- 15a2
- 30a2
- a1+30a2
- 6a2
- zero
- 8 m/s2
- 20 m/s2
- 40 m/s2
- 1 m/s
- 2 m/s
- 8 m/s
- 4 m/s
(Given that at t=0, u=2 m/s, x=4 m)
- t33−t2+2t+4
- t33−t2+2t
- t33−t2+4
- t33−t2
- 45 m/s2
- 48 m/s2
- 40 m/s2
- 52 m/s2
A car is moving along a straight line. It's displacement (x) - time (t) graph is shown. Match the entries in column I with points on graph as in Column-II.
Column I
(p) x → negative, v → positive, a → positive (q) x → positive, v → negative, a → negative
(r) x → negative, v → negative, a → positive (s) x → positive, v → positive, a → negative.
Column II
(p - d) (q - b) (r - c) (s - a)
(p - b) (q - d) (r - c) (s - a)
(p - a) (q - c) (r - b) (s - d)
(p - c) (q - a) (r - d) (s - b)
- (m2−m1)(g−ao)(m1+m2)
- (m2−m1)(g+ao)(m1+m2)
- (m2−m1)gm1
- (m2−m1)aom1
(Given, at t=0, u=3 m/s, x=2 m)
- 2t33+3t22+3t
- 2t33+3t22
- 2t33+3t22+3t+2
- 2t33+3t22−3t
- 5 m/s2
- 7.5 m/s2
- 15 m/s2
- 7 m/s2
- go on decreasing with time.
- be independent of α and β .
- drop to zero when α=β.
- go on increasing with time.
- 1
- 2
- 1 and 3
- 1, 2 and 3
more than that in case II
- equal to that in case II
Zero
- less than that in case II