Find out numerical ratio of displacement to distance for an object (shown in figure below)
that travels along the boundary of a square field of side 100m starting from a corner to
diagonally opposite corner.
Answers
Answer:
Explanation:
1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer: Initial speed of the bus, u= 0 m/s
Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 120 s
(a) v= u + at
v= 0 + 0×1 × 120
v= 12 ms-1
(b) According to the third equation of motion, v2 - u2= 2as
s is the distance covered by the bus
(12)2 - (0)2= 2(0.1) s
s = 720 m
Speed acquired finally by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
INTEXT QUESTIONS Pg 110
2. A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of ?0.5 m s-2. Find how far the train will go before it is brought to rest.
Answer: Initial speed of the train, u= 90 km/h = 25 m/s (1km/hr = 5/18 m/s)
Final speed of the train, v = 0 (finally the train comes to rest and its velocity becomes 0)
Acceleration = - 0.5 m s-2
According to third equation of motion:
v2= u2 + 2 as
(0)2= (25)2 + 2 ( - 0.5) s
Where, s is the distance covered by the train
The train will cover a distance of 625 m before coming to rest.
3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Answer: Initial Velocity of the trolley, u= 0 cms-1
Acceleration, a= 2 cm s-2
Time, t= 3 s
It is known that final velocity, v= u + at = 0 + 2*3 cms-1
Therefore, the velocity of train after 3 seconds is 6 cms-1
4. A racing car has a uniform acceleration of 4 m s - ′2. What distance will it cover in 10 s after start?
Answer: Initial Velocity of the car, u=0 ms-1
Acceleration, a= 4 m s-2
Time, t= 10 s
We know Distance, s= ut + (1/2)at2
Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102
= 0 + (1/2) × 4× 10 × 10 m
= (1/2)× 400 m
= 200 m
5. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer: Given Initial velocity of stone, u=5 m s-1
Downward of negative Acceleration, a= 10 m s-2
we know that 2 as= v2 - u2
EXERCISE QUESTIONS Pg 110
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:
Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 * ( 22 / 7 ) * 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
So, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m
Number of rounds in 40 s =1 round
Number of rounds in 140 s =140/40
=3 ½
After taking start from position X, the athlete will be at Y after 3 ½ rounds as shown in figure