If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. As the name suggests, it’s in a particular straight line, thus it can be said that it uses only one dimension.

**Q1.**

Relative velocity of ground w.r.t. B is

Solution

There is a balloon containing a stone at a height 300 m from the ground.

(Take g=10ms^(-2))

**Q2.**

A stone is dropped from the balloon when balloon is stationary at height
300m. How long will the stone take to reach the ground?

Solution

t=√(2s/g)=√((2×300)/10)=7.75s

The displacement of a body is given by 4s=M+2Nt^4, where M and N are constants.

**Q3.**

The velocity of the body at any instant is

Solution

s=(M+2Nt^4)/4 ⇒v=ds/dt=2 Nt^3

Putting t=1 s, we get v=2 N

A body is dropped from the top of the tower and falls freely.

**Q4.**

The
distance covered by it after |

Solution

S=1/2 gn^2 ⇒S∝n^2

S=a/2 (2n-1)⇒S∝(2n-1)

V=gn⇒v∝n

A body at rest is acted upon by a constant force (it means acceleration of the body will be constant).

**Q5.**

What
is the nature of the displacement-time graph? |

Solution

Here acceleration is constant. So we can use

s=ut+1/2 at^2,s-t graph will be parabolic

A car accelerates from rest at a constant rate Î± for some time and then decelerates at a constant rate Î² to come to rest. The total time elapsed is t

**Q6.**

The
maximum velocity attained by the car is |

Solution

From A to B, applying v=u+at, we get

v_0=0+at_1⇒t_1=v_0/a

From B to C, again applying v=u+at, we get

0=v_0-bt_2⇒t_2=v_0/b

Given t_1+t_2=t ⇒v_0/Î±+v_0/Î²=t⇒v_0 Î±Î²t/(Î±+Î²)

v_0 is the maximum velocity attained

A body is moving with uniform velocity of 8 ms^(-1). When the body just crossed another body, the second one starts and moves with uniform acceleration of 4 ms^(-2)

**Q7.**

The
time after which two bodies meet will be |

Solution

Let they meet after time t, then the distance travelled by both in time t should be same

s=8t=1/2 4t^2⇒t=4 s

A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is t_1 and for the remaining 50 m is t_2

**Q8.**

Which
is correct? |

Solution

s=1/2 gt_1^2 or t_1^2=(50×2)/g=100/g or t_1=10/√g

and 100=1/2 gt^2 or t=(10√2)/√g

t_2=t-t_1=10/√g (√2-1)=0.4t_1

t_1>t_2

A body is dropped from a balloon moving up with a velocity of 4 ms^(-1) when the balloon is at a height of 120.5 m from the ground

**Q9.**

The height of the body after 5 s from the ground is (g=9.8 m/s^(-2))

s=ut+1/2 at^2=4×5-1/2×9.8×5^2

=20-122.5=-102.5 m

This shows that the body is 102.5 m below the initial position, i.e.,

height of the body =120.5-102.5=18 m

A bus starts moving with acceleration 2 ms^(-2). A cyclist 96 m behind the bus starts simultaneously towards the bus at a constant speed of 20 ms^(-1)

**Q10.**

After
what time will he be able to overtake the bus? |

Solution

Let at time t, the cyclist overtake the bus, then 96+ (distance travelled by bus in time t)= (distance travelled by cyclist in time t)

⇒1/2×2×t^2+96=20×t⇒t^2-20t+96=0

This gives t=8 s or 12 s. Hence, the cyclist will overtake the bus at 8 s