Match the first column with appropriate entries in the second and third columns and remake the table. S. No. Column 1 Column 2 Column 3 1 Negative acceleration The velocity of the object remains constant A car, initially at rest reaches a velocity of 50 km / hr in 10 seconds 2 Positive acceleration The velocity of the object decreases A vehicle is moving with a velocity of 25 m / s 3 Zero acceleration The velocity of the object increases A vehicle moving with the velocity 10 m / s , stops after 5 seconds

S. No. Column 1 Column 2 Column 3 1 Negative acceleration The velocity of the object decreases A vehicle moving with the velocity 10 m / s , stops after 5 seconds 2 Positive acceleration The velocity of the object increases A car, initially at rest reaches a velocity of 50 km / hr in 10 seconds 3 Zero acceleration The velocity of the object remains constant A vehicle is moving with a velocity of 25 m / s In positive acceleration, velocity of the object increases. In the negative acceleration velocity of the object decreases; the object may still travel in the same direction, but since the final velocity is smaller than the initial velocity, the object's acceleration is negative. Zero acceleration is where the object is moving at a constant velocity relative to an inertial reference frame.

Complete the following table. u ( m / s ) a ( m / s 2 ) t ( sec ) v = u + at ( m / s ) 2 4 3 - - 5 2 20 u ( m / s ) a ( m / s 2 ) t ( sec ) s = ut + 1 2 at 2 ( m ) 5 12 3 - 7 - 4 92 u ( m / s ) a ( m / s 2 ) s ( m ) v 2 = u 2 + 2 as ( m / s ) 2 4 3 - 8 - 5 8 . 4 10

We can solve these by putting the values in the equation of motion. u ( m / s ) a ( m / s 2 ) t ( sec ) v = u + at ( m / s ) 2 4 3 14 10 5 2 20 For the first row, we have to determine the final velocity. v = u + a t ⇒ v = 2 + 4 × 3 = 14 m / s For the second row, we have to determine initial velocity. 20 = u + 5 × 2 ⇒ u = 20 - 10 = 10 m / s u ( m / s ) a ( m / s 2 ) t ( sec ) s = ut + 1 2 at 2 ( m ) 5 12 3 69 7 8 4 92 For the first row, we have to determine the displacement. s = u t + 1 2 a t 2 = 5 × 3 + 1 2 × 12 × ( 3 ) 2 = 69 m For the second row, we have to determine the acceleration. 92 = 7 × 4 + 1 2 × a × ( 4 ) 2 = 28 + 8 a ⇒ a = 64 8 = 8 m / s 2 u ( m / s ) a ( m / s 2 ) s ( m ) v 2 = u 2 + 2 as ( m / s ) 2 4 3 - 1 . 33 8 - 4 , 4 5 8 . 4 100 For the first row, we have to determine the displacement. v 2 = u 2 + 2 a s ⇒ 8 = ( 4 ) 2 + 2 × 3 × s ⇒ s = - 4 3 m = - 1 . 33 m For the second row, we have to determine the initial velocity. 100 = u 2 + 2 × 5 × 8 . 4 ⇒ u 2 = 16 ⇒ u = - 4 ms - 1 , 4 ms - 1

E M B I B E

Science and Technology Standard Nine>Chapter 1 - Laws of Motion>Can you recall?>Q 1

EASY

9th Maharashtra Board

IMPORTANT

Earn 100

What are vectors and scalars? Which of the quantities- distance, speed, velocity, time and displacement are scalars and which are vectors?

Important Questions on Laws of Motion

EASY

9th Maharashtra Board

IMPORTANT

Science and Technology Standard Nine>Chapter 1 - Laws of Motion>Use your brain power

Why is there a thick bed of sand for a high jumper to fall on after his jump?

EASY

9th Maharashtra Board

IMPORTANT

Science and Technology Standard Nine>Chapter 1 - Laws of Motion>Exercise>Q 1

Match the first column with appropriate entries in the second and third columns and remake the table.

S. No.	Column 1	Column 2	Column 3
1	Negative acceleration	The velocity of the object remains constant	A car, initially at rest reaches a velocity of $50 km / hr$ in $10$ seconds
2	Positive acceleration	The velocity of the object decreases	A vehicle is moving with a velocity of $25 m / s$
3	Zero acceleration	The velocity of the object increases	A vehicle moving with the velocity $10 m / s$ , stops after $5$ seconds