NCERT Solutions Class 9 Science Chapter 8

Q.1 An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Ans-

\begin{array}{l}\text{Given, diameter of circular track, d = 200 m}\\ \text{time required to cover the distance of circular}\\ \text{track = 40 s}\\ \text{Now, radius of the track, r =}\frac{d}{2}\text{=}\frac{\mathrm{200\; m}}{2}\text{= 100 m}\\ \text{Circumference = 2}\pi \text{r = 2}\pi (\mathrm{100\; m})=200\pi \text{m}\\ \because \text{distance covered by athlete in 40 s =}200\pi \text{m}\\ \therefore \text{in 1 s, the distance covered by the athlete}\\ \text{=}\frac{200\pi \text{m}}{40}\text{=5}\pi \text{m}\\ \text{So, the distance covered in 2 minutes and 20 seconds,}\\ \text{i}\text{.e}\text{., 140 s is given by = (140}\times \text{5}\pi )\text{m =700}\pi \text{m}\\ \text{= 700}\times \frac{22}{7}\text{m =2200 m}\end{array} \begin{array}{l}\text{The athlete completes one round in 4}0\text{s}\\ \text{So, the time taken in 3 complete rounds = 3}\mathrm{}\times \text{40 s}=\text{12}0\text{s}\\ \text{The displacement after 3 complete rounds would be zero}\text{.}\\ Now,\text{the time left = (140 s -120 s) = 20 s}\\ In\text{20 s},\text{he moves at the opposite end of the}\\ \text{initial position hence, the displacement will be equal to the}\\ \text{diameter of the circular track i}\text{.e}\text{., 200 m}\text{.}\end{array}

Q.2 Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Ans-

(a)

\begin{array}{l}\text{Given, distance covered from A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}\text{= 2}{\text{ms}}^{\text{-1}}\\ \text{displacement between A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average velocity =}\frac{\text{Net displacement}}{\text{Total time taken}}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}{\text{= 2 ms}}^{\text{-1}}\end{array}

(b)

\begin{array}{l}\text{Total distance covered from A to C = AB +BC = 300 m + 100 m = 400}\\ \text{Total time taken = time taken from A to B + time taken from B to C =}\end{array} \begin{array}{l}\text{150 s + 60 s = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{400 m}}{\text{210 s}}\text{= 1}{\text{.90 ms}}^{\text{-1}}\\ \text{Net displacement = AB – BC = 300 m – 100 m = 200 m}\\ \text{time taken = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{200 m}}{\text{210 s}}{\text{= 0.952 ms}}^{\text{-1}}\end{array}

Q.3 Abdul, while driving to school, computes the average speed for his trip to be 20 kmh⁻¹. On his return trip along the same route, there is less traffic and the average speed is 30 km h⁻¹. What is the average speed for Abdul’s trip?

Ans-

\begin{array}{l}\text{While going to school:}\\ {\text{Given, Average speed = 20 kmh}}^{\text{-1}}\\ \text{Let, the distance travelled to reach school = d}\\ {\text{time taken = t}}_{\text{1}}\\ \text{Average speed =}\frac{\text{total distance}}{\text{total time taken}}\\ \therefore \text{20 =}\frac{\text{d}}{{\text{t}}_{\text{1}}}\\ {\text{or, t}}_{\text{1}}\text{=}\frac{\text{d}}{\text{20}}\\ \text{While returning from school:}\\ {\text{Given, average speed = 40 kmh}}^{\text{-1}}\\ \text{Let, the distance travelled while returning from school = d}\\ {\text{time taken = t}}_{\text{2}}\\ \therefore \text{40 =}\frac{\text{d}}{{\text{t}}_{\text{2}}}\\ {\text{or, t}}_{\text{2}}\text{=}\frac{\text{d}}{\text{40}}\\ \\ \text{Now, Average speed =}\frac{\text{Total distance travelled}}{\text{total time taken}}\\ \text{=}\frac{\text{2d}}{{\text{t}}_{\text{1}}{\text{+t}}_{\text{2}}}\\ \text{=}\frac{\text{2d}}{\frac{\text{d}}{\text{20}}\text{+}\frac{\text{d}}{\text{40}}}\text{=}\frac{\text{80}}{\text{3}}\text{=26}{\text{.67 ms}}^{\text{-1}}\end{array}

Q.4 A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms⁻² for 8.0 s. How far does the boat travel during this time?

Ans-

\begin{array}{l}{\text{Given, u= 0, a= 3 ms}}^{\text{–}}{}^{\text{2}}\text{, t = 8 s}\\ \\ \text{On using second equation of motion,}\\ \text{s = ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ On\text{putting given values,}\\ \text{s = 0 +}\frac{\text{1}}{\text{2}}{\text{×3 ms}}^{-2}\text{×}{\left(\text{8 s}\right)}^{\text{2}}\text{= 96 m}\\ \text{Hence, the boat travels a distance of 96 m}\text{.}\end{array}

Q.5 A driver of a car travelling at 52 kmh⁻¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 kmh⁻¹ in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Ans-

\begin{array}{l}{1}^{st}\text{Case:}\\ {\text{Given, u = 52 kmh}}^{\text{-1}}\text{= 14}{\text{.4 ms}}^{\text{-1}}\\ Time\text{taken to stop the car, t = 5 s}\\ \text{v = 0}\\ Dis\tan ce{\text{covered, s}}_{\text{1}}\text{= area OAC}\\ \text{=}\frac{1}{2}\times \text{14}{\text{.4 ms}}^{-1}\times \text{5 s = 36 m}\\ 2^{nd}\text{Case:}\\ {\text{Given, u = 3 kmh}}^{\text{-1}}\text{= 0}{\text{.833 ms}}^{\text{-1}}\\ Time\text{taken to stop the car, t = 10 s}\\ \text{v = 0}\\ Dis\tan ce{\text{covered, s}}_{\text{2}}\text{= area ODB}\\ \text{=}\frac{1}{2}\times 0.833m{s}^{-1}\times 10s\text{= 4}\text{.15 m}\\ Therefore,\text{the distance covered in first case is greater}\\ \text{than the distance covered in second case}\text{.}\end{array}

Q.6 The given figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?

Ans-

(a)

$\begin{array}{l}\text{speed =}\frac{\text{distance}}{\text{time}}\\ \mathrm{slope}\text{of graph =}\frac{y-\text{axis}}{x-\text{axis}}=\frac{\text{distance}}{\text{time}}\\ \therefore \text{Speed = Slope of the graph}\\ \text{Slop of B is greater than A and C}\text{. Hence,}\\ \text{it is travelling the fastest}\text{.}\end{array}$

(b) As, A, B and C will never meet at a single point at same time therefore, they will never be at the same point on the road.

(c)

\begin{array}{l}\text{From graph, 7 small boxes = 4 km}\\ \therefore \text{1 box =}\frac{4}{7}\text{km}\\ \text{Initially, C is 4 blocks away from the}\\ \text{origin}\text{.}\\ \therefore \text{Intial distance of object C from origin =}\frac{16}{7}\text{km}\end{array}

\begin{array}{l}\text{B passes A at point P. Object C is at point Q when B passes A.}\\ \text{Distance of point Q from origin}\\ \text{= 8 km}\\ \text{Distance covered = 8 km –}\frac{\mathrm{16\; km}}{7}\text{=}\frac{\mathrm{56\; km}-\mathrm{16\; km}}{7}\text{=}\frac{\mathrm{40\; km}}{7}\\ \text{= 5}\text{.714 km}\\ \text{Therefore, C has travelled a distance of 5}\text{.714 km}\\ \text{when B passes A}\text{.}\end{array}

\begin{array}{l}\text{(d)}\end{array}

\begin{array}{l}\text{B passes C at point R.}\end{array} \begin{array}{l}\text{Distance travelled by B at the time it passes C}\\ \text{= 9 boxes =}\frac{4}{7}\times 9.25=\frac{36}{7}=5.28\text{km}\\ \text{Therefore, B has travelled a distance of 5}\text{.28 km when}\\ \text{B passes A}\text{.}\end{array}

Q.7 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms⁻², with what velocity will it strike the ground? After what time will it strike the ground?

Ans-

\begin{array}{l}Given,{\text{s = 20 m, a = 10 ms}}^{-2},\text{u = 0}\\ \text{v = ?}\\ \text{On using third equation of motion,}\\ {\text{v}}^2={\text{u}}^2+2as\\ \text{= 0 + 2}\times \text{10}\times 20=400\\ or\text{v =}\sqrt{400}=20{\text{ms}}^{-1}\\ Now,\text{using first equation of motion}\\ \text{v = u + at}\\ \therefore \text{20 = 0 + 10}\times \text{t}\end{array}

⇒ t = 2 s.

Q.8 The speed-time graph for a car is shown here.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Ans-

(a)

Distance travelled by the car the first 4s = shaded area = 12 × 4s × 6ms^-1 = 12cm

(b)

The part of the graph from 6 s to 10 s represents uniform motion of the car.

Q.10 An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Ans-

\begin{array}{l}\text{Given, radius of circular path, r = 42250 km = 42250 km ×}\frac{1000\text{m}}{1\text{km}}\\ \text{= 4}{\text{.225×10}}^{\text{7}}\text{m}\\ \text{time period, T = 24 hours = 24 hr ×}\frac{\text{60 min}}{\text{1 hr}}\text{×}\frac{\text{60 s}}{\text{1 min}}\text{= 86400 s}\\ \text{orbital velocity of satellite, v = ?}\\ \text{On using the relation,}\\ \text{v =}\frac{\text{2πr}}{\text{T}}\\ \text{On putting the given values,}\\ \text{=}\frac{\text{2×3}\text{.14×}\left(\text{4}{\text{.225×10}}^{\text{7}}\text{m}\right)}{\text{86400 s}}\text{= 3070}{\text{.94 ms}}^{\text{-1}}\end{array}