## A body sliding down on a smooth inclined plane slides down 1/4 th distance in 2 sec…

Q: A body sliding down on a smooth inclined plane slides down 1/4 th distance in 2 sec. It will slide down the complete plane in

(a) 4 sec

(b) 2 sec

(c) 5 sec

(d) 3 sec

Ans:(a)

Sol: As u = 0, and a is constant

Applying $\displaystyle S = u t + \frac{1}{2}a t^2$

In first case ,

$\displaystyle S/4 = 0 + \frac{1}{2}a 2^2$ ..(i)

In second case ,

$\displaystyle S = 0 + \frac{1}{2}a t^2$ ..(ii)

Dividing (ii) by (i), we get

$\displaystyle 4 = \frac{t^2}{4}$

t = 4 sec

## A particle moving along x-axis has acceleration f, at time t, given by f = f0 (1-t/T), where f0 and T are constants…

Q: A particle moving along x-axis has acceleration f, at time t, given by $f = f_0 (1-\frac{t}{T})$ , where f0 and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (vx) is

(a) f0T

(b) (1/2) f0T2

(c) f0T2

(d) (1/2) f0T

Ans: (d)
Sol: Putting f = 0

$\displaystyle f_0(1-\frac{t}{T}) = 0$

Hence , t = T

$\displaystyle \int_{0}^{v}dv = \int_{0}^{T}f dt$

$\displaystyle v = \int_{0}^{T}f_0(1-\frac{t}{T}) dt$

## A particle moves in a straight line with a constant acceleration. It changes its velocity from…

Q: A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms–1 to 20 ms–1 while passing through a distance 135 m in t sec. The value of t is

(a) 10

(b) 1.8

(c) 12

(d) 9

Ans: (d)
Sol:
Acceleration, a = (20-10)/t = 10/t
Using ,
$\displaystyle v^2 = u^2 + 2 a s$

$\displaystyle 20^2 = 10^2 + 2 (10/t) \times 135$

## A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in 100 s…

Q: A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?

(a) 20 ms–1

(b) 40 ms–1

(c) 25 ms–1

(d) 10 ms–1

Ans: (a)

Sol:Let speed of scooterist = v

Distance traveled by bus in 100 s is
= 10 x 100 = 1000 m

To overtake the bus , distance traveled by scooterist = 1000 + 1000 = 2000m

To overtake ,

2000 = 100 v

v = 20 m/s

## A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s1…

Q: A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s1 and that covered in the first 20 s is s2, then

(a) s2 = 2 s1

(b) s2 = 3 s1

(c) s2 = 4 s1

(d) s2 = s1

Ans: (c)

Sol:

$\displaystyle s_1 = \frac{1}{2}a (10)^2$

$\displaystyle s_2 = \frac{1}{2}a (20)^2$

On dividing ,

$\displaystyle \frac{s_1}{s_2} = \frac{1}{4}$