A body released from the top of a tower falls through half the height of the tower in 2 s. In what time shall the body fall through the height of the tower ?

4 s
3.26 s
3.48 s
2.828 s

Solution

For constant acceleration and zero initial velocity

$h\alpha t^{2}$

$\frac{h_{1}}{h_{2}}=\frac{t_{1}^{2}}{t_{2}^{2}}\Rightarrow t_{2}=\sqrt{\frac{h_{2}}{h_{1}}}t_{1}=\sqrt{2}\times t_{1}=\sqrt{2}\times 2\: s$

An electron starting from rest has a velocity that increases linearly with time i.e. v = kt where k=2 ms^-2. The distance covered in the first 3 second is

9 m
16 m
27 m
36 m

Solution

$\frac{ds}{dt}=kt\Rightarrow s=\frac{1}{2}kt^{2}=\frac{1}{2}\times 2\times 3\times 3= 9 m.$

A car travels from A to B at a speed of 20 km h^-1 and return sat as peed of 30 km h^-1. The average speed of the car for the whole journey is

5 km h^-1
24 km h^-1
25 km h^-1
50 km h^-1

Solution

Average velocity= $\frac{2\times 20\times 30}{20+30}=24kmh^{-1}$

A man leaves his house for a cycle ride. He comes back to his house after half-an-hour after covering a distance of one km. What is his average velocity for the ride ?

zero
2 km h^-1
10 km s^-1
11km s^-1

Solution

Since displacement is zero.

A boy moving with a velocity of 20 km h^-1 along a straight line joining two stationary objects. According to him both objects

move in the same direction with the same speed of 20 km h^-1
move in different direction with the same speed of 20 km h^-1
move towards him
remain stationary

Solution

Use VAB=VA-VB.

A food packet is released from a helicopter rising steadily at the speed of 2 m/sec. After 2 seconds the velocity of the packet is(g =10 m/sec²)

22 m/sec
20 m/sec
18 m/sec
none of these

Solution

he food packet has an initial velocity of 2 m/sec in upward direction, therefore
v= –u + gt or v= –2 + 10 × 2 = 18 m /sec.

A ball released from a height falls 5 m in one second. In 4 seconds it falls through

20 m
1.25 m
40 m
80 m

Solution

Since S = ut +½ gt²
where u is initial velocity & a is acceleration.
In this case u = 0 & a =g
so distance travelled in 4 sec is,
S = ½ × 10 × 16 = 80m

A particle covers half of the circle of radius r. Then the displacement and distance of the particle are respectively

2πr,0
2r,πr
^π⁄₂
πr,r

Solution

When a particle cover half of circle of radius r, then displacement is AB = 2r & distance = half of circumference of circle = πr

A person travels along a straight road for the first half time with a velocity v1 and the second half time with a velocity v2. Then the mean velocity v is given by

$\overline{V}=\frac{V_{1}+V_{2}}{2}$
$\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}$
$\overline{v}=\sqrt{v_{1}v_{2}}$
$\overline{v}=\sqrt{\frac{V_{2}}{V_{1}}}$

Solution

Let for the first half time t, the person travels a distance

s₁. Hence $v_{1}=\frac{s_{1}}{t}\: or\: s_{1}=v_{1}t$

For second half time, $v_{2}=\frac{s_{2}}{t}\: or\: s_{2}=v_{2}\: t$

Now, $\overline{v}=\frac{Total\: displacement}{Total\: time}=\frac{s_{1}+s_{2}}{2t}$

$=\frac{v_{1}t+v_{2}t}{2t}=\frac{v_{1}+v_{2}}{2}$

The displacement-time graphs of two particles A and B are straight lines making angles of respectively 30º and 60º with the time axis.If the velocity of A is vA and that of B is vB, the value of vA/vB is

1/2
$1/\sqrt{3}$
$\sqrt{3}$
1/3

Solution

Va= tan30º and Vb= tan60º

$∴\frac{V_{a}}{V_{b}}=\frac{tan30^{\circ}}{tan60^{\circ}}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}$

The study of motion, without consideration of its cause is studied in

statistics
kinematics
mechanics
modern physics

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