Question 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?

**Answer:** Here we have, Diameter = 200 m, therefore, radius = 200m/2 = 100 m

Time of one rotation = 40s

Time after 2m20s = 2 x 60s + 20s = 140s

Distance after 140 s = ?

Displacement after 140s =?

We know that, velocity along a circular path `=text{Circumference}/text{time}`

`=>v= (2pir)/(40s)`

`=>v=(2xx3.14xx100m)/(40s)`

`=>v=(628m)/(40s)`

`=>v= 15.7m//s`

(a) Distance after 140s

We know that,distance=velocity ×time

⇒ distance=15.7m/s ×140 s = 2198 m

(b) Displacement after 2 m 20 s i.e. in 140 s

Since,rotatin in 40 s=1

Therefore, rotation in `1s=1/40`

∴ rotation in `140s = 1/40 xx140 = 3.5`

Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m.

Therefore, Distance covered in 2 m 20 s = 2198 m

And, displacement after 2 m 20 s = 200m

Question 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?

**Answer:**

Here we have, Distance from point A to B = 300 m

Time taken = 2 minute 30 second = 2 × 60 + 30 s = 150 s

Distance from point B to C = 100 m

Time taken = 1 minute = 60 s

(a) Average speed and velocity from point A to B

We know that average speed `=text{Total distance}/text{time taken}`

⇒ Average speed `=(300m)/(150s)=2\ m//s`

Therefore, velocity `=2\ m//s` east

(b) Average speed and velocity from A to C

We know that average speed `=text{Total distance}/text{time taken}`

⇒ Average speed `=(400m)/(210s)=1.9\ m//s`

Average velocity `=(200)/(210)=0.95` m/s

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

**Answer:** Strategy: We need to calculate the time taken in each of the trip. After that, we can calculate the average speed.

Let the distance of the school = s km

Let time to reach the school in first trip = t_{1}

Let time to reach the school in second trip = t_{2}

We know that average speed `=text{Total distance}/text{time taken}`

∴ Average speed in first trip `=s/t_1`

`=>20\ km//h = s/t_1`

`=>t_1 = s/20 h`

∴ Average speed in second trip `=s/t_2`

`=>30\ km\\h = s/t_2`

`=>t_2 = s/30 h`

Now total time `(t_1+t_2)=s/20 +s/30`

`=>(t_1+t_2) = (3s+2s)/60 h`

`=>(t_1+t_2) =(5s)/60 h=s/12 h`

Now, Average in both of the trips `=text{Total distance covered}/text{Total time taken}`

`= (2s)/(s//12) \ km//h`

`=(2sxx12)/s \ km//h =24\ km//h`

Therefore, average speed of Adbul = 24 km/h

Question 4: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s^{2} for 8.0 s. How far does the boat travel during this time?

**Answer:** Here we have,

Initial velocity (u) = 0

Acceleration (a) = 3.0m/s^{2}

Time = 8 s

Therefore, distance (s) covered =?

We know that, `s= ut+1/2 at^2`

`=>s= 0xx8+1/2\ 3m//s^2xx(8s)^2`

`=> s = 1/2 xx 3 xx 64m`

`=>s= 3 xx32\ m`

`=>s = 96\ m`

Therefore, boat travel a distance of 96 m in the given time.

Question 5: A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h 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?

**Answer:** Given for first driver,

Intital velocity, `u=52\ km h^(-1)`

`=(52 xx1000m)/(60xx60s) =14.4\ m\ s^(-1)`

Time, `t = 5s`

Final velocity `v=0` (Since car stops)

Therefore, distance, `s=` ?

Given for second driver, `u = 3\ km\ h^(-1)`

`= (3000m)/(60xx60s)=9.4\ ms^(-1)` `

Time, `t = 10s`

Final velocity `v=0`

In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.

Distance is calculated by the area under the slope of the graph.

Thus, distance covered by 1^{st} car = Area of `triangle OAD`

⇒ Distance `s = 1/2 xx OD xx OA`

`=>s = 1/2 xx 14.4\ m//s xx 5s`

`=>s = 7.2\ m//s xx 5s = 36m`

Thus, distance covered by 2^{nd} car = Area of `triangle OBC`

⇒ Distance, `s = 1/2 xx OC xx OB`

`=> s = 1/2 xx 9.4\ m//s xx 10 s`

`=> s = 4.7\ m//s xx 10 s = 47 m`

Therefore, second car travelled farther

Copyright © excellup 2014