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.**From a high tower, at time t=0, one stone is dropped from rest and simultaneously another stone is projected vertically up with an initial velocity. The graph of distance S between the two stones plotted against time t will be

**Q2.**The displacement x of a particle moving in one dimension under the action of a constant force is related to time t by the equation t=√x+3, where x is in meters and t is in seconds. Find the displacement of the particle when its velocity is zero

Solution

t=√x+3

Differentiating will respect to t, we get

1=1/(2√x) dx/dt+0 or dx/dt=2√x

When velocity is zero, 2√x=0 or x=0

**Q3.**The x and y coordinates of a particle at any time t are given by: x=7t+4t^2 and y=5t, where x and y are in meters and t in seconds. The acceleration of the particle at 5 s is

Solution

a_x=(d^2 x)/(dt^2 )=8 and a_y=(d^2 y)/(dt^2 )=0

Hence, net acceleration is √(a_x^2+a_y^2 )=8 ms^(-2)

**Q4.**A particle starts from the origin with a velocity of 10 ms^(-1) and moves with a constant acceleration till the velocity increases to 50 ms^(-1). At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

Solution

2ax=(50)^2-(10)^2 and 2(-a)(-x)=v^2-(50)^2

This given v^2-(50)^2=(10)^2 i.e., v=70 ms^(-1)

**Q5.**Which graph represents uniform motion?

Uniform motion involves equal distances covered in equal time intervals or the velocity is constant.

**Q6.**

A stone is dropped from rising balloon at a height of 76 m above the ground and reaches the ground in 6 s. What was the velocity of the balloon when the stone was dropped? Take g=10 ms^(-2) |

Solution

S=ut+1/2 at^2

⇒ -76=u×6-1/2×10×(6)^2⇒u=52/3 ms^(-1)

**Q7.**

The distance travelled by a particle in a straight line motion is directly proportional to t^(1/2), where t= time elapsed. What is the nature of motion?

Solution

s=kt^(1/2) ⇒a=(d^2 s)/(dt^2 )=-1/4 kt^(-3/2)

As t increases, retardation decreases

**Q8.**From the velocity-time graph, given in Fig of a particle moving in a straight line, one can conclude that

Solution

Displacement in 12 s = area under v-t graph

=1/2×(12+5)4=34 m

Vav=Displacement/Time=34/12=17/6 ms^(-1)

Hence (a) is incorrect; (b) is incorrect because during first 3 seconds, velocity increases from 0 to 4 ms^(-1) option; (c) is incorrect, because in part AB velocity is constant

**Q9.**

Plot the acceleration-time graph of the velocity-time graph given in Fig |

Solution

For 0 to 5 s, acceleration is positive, for 5 to 15 s acceleration is negative, for 15 to 20 s acceleration is positive.

**Q10.**A stone is dropped from a certain height which can reach the ground in 5 s. It is stopped after 3 s of its fall and then it is again released. The total time taken by the stone to reach the ground will be

Solution

Here h=1/2×10×(5)^2=125 m

In 3 s it falls through: h_1=1/2×10×(3)^2=45 m

Rest 80 m is covered in 4 s. Hence, total time taken is 3 s+4 s=7 s