Numericlas of Ex.2(A) Q8-Q16 based on chapter 2 (motion in One dimension) of class 9

 8. A car moving on a straight path covers a distance of 1 kilometre due east in 100 seconds. What is a)the speed and b)the velocity of car?

Ans. Distance covered=1km 

distance covered=1×1km

               =1×1000m  (1km=1000m)

                =1000m

Time=100s

 speed= distance covered/ time taken

Speed=1000m/100s

           =10m/s

Velocity=Speed in a specific direction

Velocity=10m/s in east direction.

9. A body starts from rest and acquires a velocity 10 metre per second in 2 seconds. Find the acceleration?

Ans. Starts from rest,u=0

Final velocity after 2 seconds,v=10ms-¹

Time,t=2s

Acceleration=?

Acceleration= final velocity- initial velocity / time taken

a = v-u / t

a= 10ms-¹ - 0ms-¹ / 2s

a=10ms-¹ / 2s

a = 5ms-².

10. A car starting from rest acquires a velocity 180 metre per second in 0.05 hours. Find the acceleration?

Ans. Starts from rest,u=0ms-¹

Acquires a velocity,v=180ms-¹

Time taken for it,t=0.05h

               t=0.05×1h   (1h=60×60s)

               t=0.05×60×60s

               t=0.05×100×6×6 s

                t=5×36 s

                 t=180s

a = (v - u) / t

a= (180ms-¹ - 0ms-¹) / 180s

a=180ms-¹ / 180s

a=1ms-²

11. A body is moving vertically upwards. Its velocity changes at a constant rate from 50 metre per second to 20 metre per second in 3 seconds. What is its acceleration?

Ans.u=50 ms-¹

v=20ms-¹

t=3s

a=?

a=v- u / t

a=(20ms-² - 50ms-¹) / 3s

a=-30ms-¹ / 3s

a= - 10ms-².

Negative sign indicates that velocity decreases with time. Negative acceleration is known as retardation. Retardation is 10 ms-²

12. A toy car initially moving with a uniform velocity of 18 kilometre per hour comes to a stop in 2 seconds. Find the retardation for the car in SI units?

Ans. u=18kmh-¹

u=18×1kmh-¹

u=18×5/18  ms-¹

u=5ms-¹.

Car stops, v=0ms-¹

t= 2s.

a=?

a = v- u /t

a = 0 - 5ms-¹ / 2s

a= - 5 ms-¹ / 2s

a= - 2.5 ms-².

Multiplying both sides by -1.

-1 × a = -1 × -2.5ms-²

-a = 2.5 ms-²

Retardation= negative acceleration. 

Retardation= - a

Retardation = 2.5ms-².

13. A car accelerates at a rate of 5 metre per second square. Find the increase in its velocity in 2 seconds?

Ans.a= 5ms-²

t=2s

Increase in velocity=?

Increase in velocity= final velocity -initial velocity

Increase in velocity= v - u

(v- u ) / t  = a

v - u = a × t

v - u = 5ms-² × 2s

v - u = 10ms-¹

So increase in velocity is 10 metre per second.

14. A car is is moving with a velocity 20 metre per second. The brakes are are applied to retard at a rate of 2 metre per second square. What will be the velocity after 5 seconds of applying the brakes?

Ans. A car is moving with a velocity of

 20 metre per second.

u= 20ms-¹

v=?

a= -2ms-² (negative sign because of retardation)

t=5s

(v - u) / t  = a

v - u = a × t

v = at + u

v= -2ms-² ×5s + 20ms-¹

v= -10ms-¹ + 20 ms-¹

v = 10 ms-¹.

15. A bicycle moving with velocity 5.0 metre per second accelerates for 5 second at a rate of 2 metre per second square. What will be its final velocity?

Ans.u=5.0 ms-¹

t=5s

a=2ms-²

v=?

(v - u )/ t = a

v- u = a×t

v = at + u

v= 2ms-² × 5s  + 5.0ms-¹

v=10ms-¹ + 5.0ms-¹

v= 15ms-¹.

16. A car is is moving in a straight line with speed 18 kilometre per hour. It is stopped in 5 seconds by applying the brakes. Find: a) the speed of car in metre per second, b) the retardation c) the speed of car after 2 seconds of applying the brakes.

Ans.u=18kmh-¹

u = 18×1kmh-¹

    = 18 × 5/18 ms-¹ (1kmh-¹=5/18 ms-¹)

    =5ms-¹.

t= 5s

Car stopped .It means final velocity is zero.

v= 0ms-¹.

a = (v - u )/t

a=(0ms-¹ - 5ms-¹) / 5s

a=-5ms-¹ / 5s

a= - 1 ms-².

Retardation= -a

                      =-(-1ms-²)

                       = 1ms-²

The speed of car after 2 second of applying the brakes, t= 2s

u=5ms-¹

v=?

a=-1ms-²

(v - u)/t = a

v - u = a×t

v= at + u

v= -1ms-²×2s + 5ms-¹

v=-2ms-¹ + 5ms-¹

v= 3ms-¹.