# Class 9 Motion Important Questions : Solutions

Important Questions Notes NCERT Solutions**Motion Important Questions** **: Solution**

**Define uniform circular motion and give example of it. Why is it called accelerated motion?****Sol:-**Uniform circular motion can be described as the motion of an object in a circular path at a constant speed.

Example:- Motion of an electron around the nucleus in a circular orbit, motion of the tip of second’s hand of a watch etc.

Circular motion is always an accelerated motion because the direction of motion changes at every point on the circular path.**A car moves with a speed of 30 km/h for half an hour, 25 km/h for one hour and 40 km/h for two hours. Calculate the average speed of the car.****Sol:-**As we know that,

Average speed =Total distance travelledTotal time taken

Now,

Time = 0.5 + 1 + 2 = 3.5 hr

Distance = (30 x 0.5) + (25 x 1) + (40 x 2) = 15 + 25 + 80 = 120 Km

Hence ,

Average speed = ≈ 34.3 Km/hr**Derive the equation for position – time relation by graphical method.****Sol:-**

Consider an object moving with initial velocity ‘u’ and uniform acceleration ‘a’. After time ‘t’, let its final velocity becomes ‘v’.

From graph,

OD = u, OC = BE = v, OE = t

Distance travelled = Area of trapezium ODBE

⇒ S = ½ ( sum of parallel sides ) x height

⇒ S = ½ (OD + BE) x OE

⇒ S = ½ (u + v) x t

⇒ S = ½ (u + u + at) x t*[∵ v = u + at]*

⇒ S = ½ (2u + at) x t

⇒ S = ut + ½ at^{2}**A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s**^{-2}, with what velocity will it strike the ground? After what time will it strike the ground?**Sol:-**Given,

u= 0 m/s, g = 10 m/s^{2}, h = 20 m

Using, v^{2}– u^{2}= 2aS

⇒ v^{2}– 0^{2}= 2 x 10 x 20

⇒ v^{2}= 400

⇒ v =

⇒ v = 20 m/s

Using, v = u + at

⇒ 20 = 0 + 10 x t

⇒ 20 = 10t

⇒ t = 2 sec

Hence, the ball will strike the ground with the velocity of 20 m/s after 2 sec.**A train starts its journey from station P; accelerates at the rate of 2 m s**^{-2}, and reaches its maximum speed in 10 s. It maintains this speed for 30 min and retards uniformly to rest at the station Q after the next 20 s. Calculate:**the maximum speed of the train.****retardation.****the distance between stations P and Q.****Sol:-**(**1**) Given,

u = 0 m/s, a = 2 m/s^{2}, t = 10 s

Using, v = u + at

⇒ v = 0 + 2 x 10

⇒ v = 20 m/s

(**2**) Given,

u = 20 m/s, v = 0 m/s, t = 20 s

Using, a = = = -1 m/s^{2}

(**3**) S_{1}= ut + ½ a t^{2}= 0 x 10 + ½ x 2 x 10^{2}= 100 m

S_{2}= vt = 20 x 30 x 60 = 36000 m

S_{3}= ut + ½ at^{2}= 20 x 20 + ½ x (-1) x 20^{2}= 200 m

Total distance = S_{1}+ S_{2}+ S_{3}= 100 + 36000 + 200 = 36300 m = 36.3 km

**A boy runs for 20 min at a uniform speed of 18 km/h. At what speed should he run for the next 40 min so that his average speed becomes 24 km/h?****Sol:-**Given,

t_{1}= 20 min = ⅓ hr and t_{2}= 40 min = ⅔ hr

v_{1}= 18 km/hr

Average speed = 24 km/hr

Total time = t_{1}+ t_{2}= 20 + 40 = 60 min = 1 hr

Distance covered in 1 hr = Speed x time = 24×1 = 24 km

Distance covered in 20 min (⅓ hr) = v_{1}x t_{1}= 18 x ⅓ = 6 km

Distance covered in 40 min (⅔ hr) = Distance covered in 1 hr – Distance covered in 20 min (⅓ hr) = 24 – 6 = 18 km

Speed needed for the next 40 min = v_{2}= = = 18 x = 27km/hr

Hence, he should run at the speed of 27km/hr for the next 40 min so that his average speed becomes 24 km/h.**A train accelerated from 10 km/h to 40 km/h in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight.****Sol:-**Given,

u = 10 km/hr, v = 40 km/hr, t = 2 min = hr

Using, v = u + at

a = = = ( 40 – 10 ) x 30 = 900 km/hr^{2}

So, the distance covered = S = ut + 1/2at^{2}

⇒ S = 10 x 1/30 + ½ x 900 x (1/30)^{2}

⇒ S = ⅓ + 450 x 1/900 = ⅓ + ½ = ⅚ km

Hence, distance covered in by the train 2 mins = ⅚ km.**A train starts from rest and accelerates uniformly at the rate of 5 m s**^{-2}**for 5 sec. Calculate the velocity of train in 5 sec.****Sol:-**Given,

u = 0 m/s, a = 5 m/s^{2}, t = 5 s

Using, v = u + at

⇒ v = 0 + 5 x 5

⇒ v = 25 m/s

Hence, the velocity of train in 5 s is 25 m/s.**A bullet leaves a rifle with a muzzle velocity of 1042 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 1.680 m. Determine the acceleration of the bullet (assume a uniform acceleration).****Sol:-**Given,

u = 0 m/s, v = 1042 m/s and S = 1.680 m

Using, v^{2}– u^{2}= 2aS

⇒ 1042^{2}– 0^{2}= 2 x a x 1.680

⇒ a = 323144.048 m/s^{2}

Hence, the acceleration of the bullet is 323144.048 m/s^{2}.**A bike riding at 22.4 m/s skids to come to a halt in 2.55 s. Find the skidding distance of the bike.****Sol:-**Given,

u = 22.4 m/s, v = 0 m/s, t = 2.55 s

Using, v = u + at

⇒ a =

⇒ a =**≈**-8.7 m/s^{2}

Now,

Using, S = ut + ½ a t^{2}

⇒ S = 22.4 x 2.55 + ½ x -8.7 x 2.55^{2}

⇒ S = 57.12 + (-28.28) = 28.84 m

Hence, the skidding distance of the bike is 28.84 m.**A race of scooter is seen accelerating uniformly from 18.5 m/s to 46.1 m/s in 2.47 sec. Determine the acceleration of the scooter and the distance traveled.****Sol:-**Given,

u = 18.5 m/s, v = 46.1 m/s, t = 2.47 s

Using, a =

⇒ a = = 27.6 / 2.47**≈**11.17 m/s^{2}

Using, 2aS = v^{2}– u^{2}

⇒ 2 x 11.17 x S = 46.1^{2}– 18.5^{2}

⇒ 22.34 x S = 2125.21 – 342.25

⇒ S = 1782.96 / 22.34

⇒ S = 79.81 m

Hence, the acceleration is 11.17 m/s^{2}and distance traveled is 79.81 m**A car is travelling with a speed of 36 km/h. The driver applied the brakes and retards the car uniformly. The car is stopped in 5 sec. Find (i) the acceleration of car and (ii) distance before it stops after applying brakes?****Sol:-**Given,

u = 36 Km/h = 36 x 5 / 18 = 10 m/s

v = 0 m/s

t = 5 s

Using, a =

⇒ a = = -2 m/s^{2}

Using, S = ut + ½ at^{2}

⇒ S = 10 x 5 + ½ x (-2) x 5^{2}

⇒ S = 50 – 25 = 25 m

Hence, (i) The acceleration of car is -2 m/s^{2}and (ii) distance before it stops after applying brakes is 25 m.**Robbers in a car travelling at 20 m/s pass a policeman on a motorcycle at rest. The policeman immediately starts chasing the robbers. The police man accelerates at 3m/s**^{2}for 12 s and thereafter travels at a constant velocity. calculate the distance covered by the policeman before he overtakes the car.**Sol:-**Let police overtake the robbers in time ‘t’

So, Distance covered by robbers in time ‘t’ = speed x time

= 20t …..(i)**For police**

u = 0 m/s, t = 12 s, a = 3 m/s^{2}Using,

⇒ v = 0 + 3 x 12

⇒ v = 36 m/s

Distance covered in 12 s = S = ut + ½ at^{2}

⇒ S = 0 x 12 + ½ x 3 x 12^{2}

⇒ S = 216 m

Let total time taken to overtake be ‘t’ sec.**∴**Remaining time = (t – 12) sec

So, Distance covered = speed x time = 36(t – 12) m

Now,

Distance covered by robbers = Distance covered by police

⇒ 20t = 216 + 36 x (t – 12)

⇒ 20t = 216 + 36t – 432

⇒ 36t – 20t = 432 – 216

⇒ 16t = 216

⇒ t = 13.5 s

So, Distance covered by robbers = 20t = 20 x 13.5 = 270 m

Hence, the distance covered by the policeman before he overtakes the car is 270 m.**The variation of the velocity of a particle moving along a straight line is illustrated in the graph given below. Find the distance covered by the particle in 4 seconds**.**Sol:-**

Distance covered by the particle in 4 sec

= Area under the graph

= Area of trapezium OABG + Area of trapezium BCFG + Area of rectangle CDEF

= ½ (AB+OG) x AH + ½ (CF+BG) x GF + DE x FE

= ½ (1+2) x 20 + ½ (10+20) x 1 + 10 x 1

= 30 + 15 + 10

= 55 m

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