The velocity of a particle at an instant is 10 m/s. After 5 sec,the velocity of the particle is 20 m/s. Find the velocity at 3 seconds before from the instant when velocity of a particle is 10m/s.

8 m/s
4 m/s
6 m/s
7 m/s

Solution

u= 10 m/s, t= 5 sec, v =20 m/s, a =?

a =\(\frac{20-10}{2}\)=2 ms^-2

From the formula v₁= u1₁+ a t,we have

10 = u₁+ 2 × 3 or u₁= 4 m/sec

A particle accelerates from rest at a constant rate for sometime and attains a velocity of 8 m/sec. Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is

32 m
16 m
4 m
None of the above

Solution

The two ends of a train moving with constant acceleration pass a certain point with velocities u and v. The velocity with which the middle point of the train passes the same point is

(u + v)/2
(u² + v²)/2
\(\sqrt{(u^{2}+v\vartheta ^{2})/2}\)
\(\sqrt{(u^{2}+v\vartheta ^{2})/2}\)

Solution

Let the length of train is s, then by third equation of motion, v²=u²+2a×s....(1)

Where v is final velocity after travelling a distance s with an acceleration a & u is initial velocity as perquestion Let velocity of middle point of train at same point is v',then

(v')²=u²+2a(s/2)....(2)

By equations (1) and (2), we get v'=\(\sqrt{\frac{v^{2}+u^{2}}{2}}\)

When the speed of a car is v, the minimum distance over which it can be stopped is s. If the speed becomes n v, what will be the minimum distance over which it can be stopped during same retardation

s/n
n s
s/n²
n²s

Solution

v² = u²+ 2 as or v²– u²= 2 a s

Maximum retardation, a = v²/2 s

When the initial velocity is n v, then the distance over which it can be stopped is given by

\(S_{n}=\frac{u_{0}^{2}}{2a}=\frac{(nv^{2})}{2(v^{2}/2s)}=n^{2}s\)

A bus starts moving with acceleration 2 m/s². A cyclist 96 m behind the bus starts simultaneously towards the bus at 20 m/s. After what time will he be able to overtake the bus?

4 sec
8 sec
12 sec
16 sec

Solution

A point moves with uniform acceleration and v₁, v₂ and v₃ denote the average velocities in t₁, t₂ and t₃ sec. Which of the following relation is correct?

(v₁-v₂):(v₂-v₃)=(t₁-t₂):(t₂+t₃)
(v₁-v₂):(v₂-v₃)=(t₁+t₂):(t₂+t₃)
(v₁-v₂):(v₂-v₃)=(t₁+t₂):(t₁-t₃)
(v₁-v₂): (v₂-v₃)=(t₁+t₂):(t₂-t₃)

Solution

A passenger travels along the straight road for half the distance with velocity v1and the remaining half distance with velocity v₂. Then average velocity is given by

v₁ v₂
v₂²/ v₁²
(v₁ + v₂)/2
2v₁ v₂ / (v₁+ v₂)

Solution

The displacement of aparticle is given by
y = a + b t + c t²– d t⁴
The initial velocity and acceleration are respectively

b, – 4d
– b, 2c
b, 2 c
2 c, –4 d

Solution

The acceleration of a particle,starting from rest, varies with time according to the relation
a=-ω² sin ω t
The displacement of this particle at a time t will be

s sin ω t
s ω cos ω t
s ω sin ω t
-¹⁄₂(s² ω sin ω t)t²

Solution

The position x of a particle varies with time (t) as x =A t²– B t³. The acceleration at time t of the particle will be equal to zero. What is the value of t?

\(\frac{2A}{3B}\)
^A⁄_B
\(\frac{A}{3B}\)
zero

Solution

Given that x = A t²– Bt³

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