# Illustrations

A particle moves along a circle of radius *R*. When it has covered quarter of the circumference, its displacement is ________.

- R
^{2} - zero

Ans (3)

Displacement of the particle is given by

(By Pythagoras theorem *AB*^{2} = *OA*^{2} + *OB*^{2})

Light takes 8.5 min to travel from the sun of the earth. The velocity of light 3 Ã— 10^{5} km s^{âˆ’1}. Then the distance of the sun from the earth is ________.

- 1.53 Ã— 10
^{8}km - 3 Ã— 10
^{8}m - 4 Ã— 10
^{8}km - 2.55 Ã— 10
^{6}km

Ans (1)

Distance of sun from earth

= Speed of light Ã— Time taken by light

= 3 Ã— 10^{5} km s^{âˆ’1} Ã— (8.5 Ã— 60) s

= 1.53 Ã— 10^{8} km

The distance travelled by a car in the first 5 s as per the given velocityâ€“time graph is ________.

- 100 m
- 25 m
- 4 m
- 0 m

Ans (1)

Displacement in 5 s

= Area of rectangle *OABC*

= *OA* Ã— *OC* = 5 Ã— 20 = 100 m

A car acquires a velocity of 72 km h^{âˆ’1 }in 10 s starting from rest. Calculate

- the acceleration
- the average velocity
- the distance travelled in this time.

Given: Initial velocity *u* = 0 m s^{â€“1}

Final velocity

Time taken *t* = 10 s

- Acceleration
*a*= 2 m s^{â€“2} - Average velocity
- Distance travelled
*s*= Average velocity Ã— Time^{â€“1}) Ã— (10 s) = 100 m^{â€“1 }Ã— 10 s + m s^{â€“2}Ã— 10 s Ã— 10 s

A stone thrown vertically upwards takes 3 s to attain the maximum height. Find the

- initial velocity of the stone
- maximum height attained by the stone (
*g*= 9.8 m s^{â€“2})

Given: Acceleration due to gravity, *g* = 9.8 m s^{â€“2}

Final velocity, *v* = 0 m s^{â€“1}

Time, *t* = 3 s

Distance = Height = ?

Here, *u* is upwards and *g* is downwards. We take *u* as +ve and *g* as âˆ’ve.

- To find the initial velocity
*v*=*u*+*at*(here*a*= âˆ’*g*)*v*=*u*âˆ’*gt**u*=*v*+*gt**= 0 m s*^{â€“1}+ 9.8 m s^{â€“2}Ã— 3 s*= 29.4 m s*^{â€“1} - To find the height
*s*=*ut*^{â€“1}Ã— 3 s) â€“ (9.8 m s^{â€“2}Ã— (3 s)^{2}^{= 88.2 m â€“ 44.1 m = 44.1 m}^{Therefore, the maximum height is 44.1 m.}