If a car at rest accelerates uniformly to a speed of 144 km/h in 20 sec., it covers a distance of

20 cm
400 m
1440 cm
2980 cm

Solution

v = [144 × 1000/(60 × 60)] m/sec.
v = u + at
or (144 × 1000)/(60 × 60) = 0 + a × 20
\(∴a-\frac{144\times 1000}{60\times 60\times 20}-2m/sec^{2}\)
now s=u t+\(\frac{1}{2}at^{2}=0+\frac{1}{2}\times 2\times (20)^{2}=400 \: m\)

If a ball is thrown vertically upwards with a velocity of 40m/s, then velocity of the ball after two seconds is :(g =10 m/sec²)

15 m/s
20 m/s
25 m/s
28 m/s

Solution

From first equation of motion v = u + a t
hereu = 40, a = g = –10, t= 2
so v = 40 – 10 × 2 = 20 m/sec

A boat takes 2 hours to travel 8 km and back instill water lake. With water velocity of 4 km h^-1, the time taken forgoing upstream of 8 km and coming back is

160 minutes
80 minutes
100 minutes
120 minutes

Solution

Velocity of boat=\(\frac{8+8}{2}=8km\: h^{-1}\)

Velocity of water 14 kmh^-1

\(t-\frac{8}{8-4}+\frac{8}{8+4}-\frac{8}{3}h-160\) minute

Figure here gives the speed-time graph for a body. The displacement travelled between t = 1.0 second and t = 7.0 second is nearest to

1.5 m
2 m
3 m
4 m

Solution

\(\frac{1}{2}(2+4)\times 2+\frac{1}{2}\times \frac{1}{2}\times 3\times 4=2m\)

A body is thrown upwards and reaches its maximum height.At that position

its acceleration is minimum
its velocity is zero and its acceleration is also zero
its velocity is zero but its acceleration is maximum
its velocity is zero and its acceleration is the acceleration due to gravity.

A body moves in a straight line along Y-axis.Its distance y(in metre) from the origin is given by y = 8t – 3t². The average speed in the time interval from t = 0 second to t = 1 second is

–4 ms^-1
zero
5 ms^-1
6 ms^-1

Solution

\(\overline{V}-\frac{(3\times 1\times 3\times 1\times 1)}{1}=5ms^{-1}\)

Two bodies of masses m₁ and m₂ fall from heights h₁ and h₂ respectively. The ratio of their velocities, when they hit the ground is

\(\frac{h_{1}}{h_{2}}\)
\(\sqrt{\frac{h_{1}}{h_{2}}}\)
\(\frac{m_{1}h_{1}}{m_{1}h_{1}}\)
\(\frac{h_{1}^{2}}{h_{2}^{2}}\)

Solution

Whena body falls through a height h, it acquires a velocity\(\sqrt{2gh}\)

The displacement of a particle is given by \(\sqrt{x}\)=t+1. Which of the following statements about its velocity is true ?

It is zero
It is constant but not zero
It increases with time
It decreases with time

Solution

\(x-t^{2}+2t+1\)

Hence,\(v-\frac{dx}{dt}-2t+2\)It increases with time.

The displacement × of a particle at the instant when its velocity is v is given by v=\(\sqrt{3x+16}\) . Its acceleration and initial velocity are

1.5 units, 4 units
3 units, 4 units
16 units, 1.6 units
16 units, 3 units

Solution

\(v=\sqrt{3x+16}\Rightarrow v^{2}-3x+16\)

\(\Rightarrow v^{2}-16-3x\)

Comparing with v²-u²=2aS,we get,u = 4 units, 2a= 3 or a = 1.5 units

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