Motion In One Dimension(11th And 12th > Physics ) Questions and Answers
The position of a particle travelling along x - axis is given by xt = t3 - 9 t2 + 6t where xt is in m and t is in second . Then
1) the particle does not come to rest at all.
2) the particle comes to rest firstly at (3 - √7 ) s and then at (3 + √7) s
3) the speed of the particle at t= 2s is 18 ms−1
4) the acceleration of the particle at t= 2s is 6 ms−2
Explanation:-
Answer: Option A. -> 2,3, 4 are correct:
A
xt = t3 - 9 t2 + 6t
v= dxtdt = (3 t2 - 18t + 6) m s−1
So, the particle comes to rest since this equation has two valid roots.
⇒ t1 = (3 - √7) s and ⇒ t2 = (3 - √7) s
Substituting the value t = 2s in the equation for velocity, we get, v= = - 18 ms−1
The acceleration of the particle is a=6t - 18 ms−2
Substituting the value t = 2s in the equation for acceleration, we get, a= = - 6 ms−2
Explanation:-
Answer: Option C. -> v = (αβ) (1 - e−βt):
C
Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity.
dvdt = α - β v ⇒ v∫1 dvα−βv = t∫0 dt
v = (αβ) (1 - e−βt)
Explanation:-
Answer: Option A. -> 2.4:
A
In direction of stream , t1 = 23+2 = 0.4 h
In upstream t2 = 23−2 = 2 h
Total Time = 2.4 h
Explanation:-
Answer: Option A. -> 105 m s−1:
A
The speed of the bullet with respect to ground will be,
vbg = vbv + vvg
=150 + 253 = 4753ms−1
Speed of thief's car is,
vtg=192 × 518 = 1603ms−1
vbt = vbg - vtg
4753 - 1603 = 105 ms−1
Explanation:-
Answer: Option C. -> 2√Rg:
C
∠ABC=900Acceleration along AB is a=gcos θDistance travelled is AB=2R CosθWe have 2R Cosθ=12×g cosθ t2⇒t=2√Rg
Explanation:-
Answer: Option D. -> At 3h4 from the ground:
D
Let the body after time t2 fall a distance x from the top. Since it starts from rest, we have
x=12gt24=gt28 .............(i)
It reaches the ground in t seconds, we have
h=12gt2 ..............(ii)
Eliminate t from (i) and (ii), we get x = h4
∴ Height of the body from the ground
= h−h4=3h4
Explanation:-
Answer: Option D. -> 24.5 m:
D
The first body would have travelled for 3s. Since both of them start from rest,the
separation between the two bodies, two seconds after the release of second body is
x=12×9.8[(3)2−(2)2]=24.5m