1.Acar cover 30km at a unoform speed of 60km/hr and the next 30km at a uniform speed of 40km/hr.Find the total time taken.

Sol:-For uniform speed,

s=vt

If the car text time t1, to cover the first 30 km

30 Km=(60km/h)×t1

or,t1=30km/60km/h=1/2=30min

Similarly, if it takes time t2 to cover the next 30 km

30km=(40km/h)×t2

or,t2=30km/40km/hr=3/4h=45min

Total time taken

t1+t2=30min+45min=75min

1h15min

Ans.1hr15min

2.A particle moving with an initial velocityof 5.0m/s is subjected to a uniform acceleration of -2.5m/s^2.Find the displacement in 4.0 s.

Sol:-the displacement is

s=ut+1/2at^2

=(5.0ms)(4.0s)+1/2(-2.5ms^2)(4.0)^2

=20m-20m=0

So, after 4.0 second the particle will be back at its initial position

Ans.0

3.A boy runs for 10 minutes at a uniform speed of 9km/hr.At what speed should he run for the next 20 minutes so that the average speed comes to 12 km/hr.

Sol:-Total time =10min+20min=30min

Total average speedis 12km/hr

using s=vt

Total distance covered in 30min=12km/hr x30min

=12km/hr x 1/2h=6km

Distance covered in the first 10 min

9km/hr x 10 min=9km/hr x 1/6hr=1.5km

Thus,he has to cover 6km-1.5km

=4.5 km in next 20 min

The speed required is 

4.5km/20min=4.5km/(20/60)h=13.5 km  

Ans. 13.5 km per hour

 4.A train accelerates from 20 km per hour to 80 km per hour in 4 minutes. How much distance does it cover in this period? Assume that the tracks are straight.

Sol:-We will first find the acceleration and the distance

At t=0,velocity is u = 20km/hr

At t=4min,=1/15hr,the velocity is v=80km/hr

using,v=u+at

a=v-u/t=80km/hr-20km/hr/(1/15)h=60km/hr^2

60km/hr^2 x 15=900 km/hr^2

The distance covered is 

x=ut+1/2at^2

= (20km/hr)(1/15hr)+1/2(900km/hr^2) x (1/15hr^2)

20/15 km + 2km= 10/3km

Ans.10/3 kilometre

5. A boy leaves his house at 9:30 a.m. for school. The school is 2 km away and classes start at 10:00 a.m. If he walks at a speed of 3 km per hour for the first kilometre, at what speed should he walk the second kilometre to reach just in time.

Sol:-Total time =30min=(0.5hr)

total distance=2km

(A to q) 1st case

distance travelled=1km

speed=3km/hr

time taken=1/3hr (distance / speed )

2nd case

distance left or have to travel=1km

time left=1/3 x 0.5 hours

=0.5/3 hours

speed = distance/time

=1/0.5/3

=1 x 3/0.5

=6km/hr

Ans. 6 km per hour

6. During an experiment,a signal from a spaceship reach the ground station in five minutes. what was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3×10^8m/s.

Sol:-Here,

t=5minutes=5 x 60seconds=300 seconds

v=3 x 10^8m/s

s=vt,we get

s=3 x 10^8 x 300  s=9 x10^10m

=9 x 10^7km

Therefore,the distance of the spaceship from the ground was 9 x 10^7 km

Ans.9×10^7 km

7. A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 110 m, calculate the speed of cyclist.

Sol:-Radius of circular track,r=110m

distance travelled by a cyclist in 2 min=circumference of circular track

               2Ï€
r

=2 x 22/7 x 110=4840/7m

time taken,t=2 minutes=2 x 60 =120 s
speed of cyclist v = s/ t=4840/7 x 120s = 5.76m/s

Ans. 5.76 metre per second

8.An object starting from rest travels 20 m in first 2 seconds and 160 m in next 4 seconds.What will be the velocity after 7 s from the start.

Sol:-1st case

here,u=0,s1=20m,t1=2s

using s1=ut1+1/2at1^2

we get,

20=0+1/2 x a x 4

or, a=10m/s^2

2nd case

s2=160m,t2=4s

s1+s2=20+160=180m

t1+t2=2+4=6s

180=1/2a x 36=18a

or,a=10m/s^2

velocity after 7s,v=u+at=0+10 x 7=70m/s

Ans. 70 metre per second

9. A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in 10 second after the start.

Sol:-s=ut+1/2at^2

s=displacement

u=0(initial velocity) starting from rest

a=4m/s^2

s=ut+1/2at^2=0 x 10+1/2 x 10 x 4^2 or, s=200m

Ans. S  200 m

10.A bus increase its speed from 20 km per hour to 50 km per hour in 10 seconds its acceleration is.

Sol:-u=20km/hr

=20 x 1000/60x60

=5.5m/s

v=50km/hr

=13.8m/s

v=u+at

13.8=5.5+10a

10a=8.3

a=0.83m/s^2

Ans. 0.83 metre per second square