Solutions to Exercises on Kinematics - Rest And Motion HC Verma's Concepts of Physics Part 1 (22-26)

 Q#22

A car traveling at 60 km/h overtakes another car traveling at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtaking and the total road distance used for the overtake.

Answer:
60 km/h = 60000/3600 m/s =50/3 m/s

42 km/h = 42000/3600 m/s = 35/3 m/s

Relative speed of faster car with respect to slower car

= 50/3 – 35/3 m/s =  5 m/s

Total distance to be covered in overtaking = Total length of both cars = 10.0 m.

So, the time taken in overtaking = 10/5 s = 2 s.

Total road distance used for the overtaking will be the distance covered by overtaking the car in this time plus the length of the car = (50/3) x 2 + 5 ≈ 38 m (It can be illustrated by the following diagram)

Q#23
A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g= 10 m/s².

Answer:
u = 50 m/s, v = 0 (at maximum height), Let Maximum height be h. The relationship is given by the equation,

v² = u² - 2gh
0 = u² - 2gh
h = u²/2g = 2500/20 m =125 m.

If t is the time to reach the maximum height, we can use the following equation,
h = ut – ½gt²
125 = 50t – ½(10)t²
t² – 10t + 25 = 0
(t – 5)² = 0
t = 5 s

Half height = s = 125/2 m,  Speed at half height v is given by the equation,

v² = u² – 2gs = 50² – 2(10)(125/2) = 1250
v = √1250 = 35.3 m/s ≈ 35 m/s.

 Q#24
 A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height of 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

Answer:
Let the time taken by the ball to reach the ground be t. We have u = -7 m/s (Taking downward direction as +ve), h = 60 m and g = 10 m/s². Using equation h = ut + ½gt²,

60 = -7t + ½(10)t²

5t² – 7t – 60 = 0  We solve this quadratic equation for t and take only the positive value

T = {7 ± √(49 + 4.5.60)}/(2 x 5)

t = 4.23 s

Q#25
A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone, (b) Find its velocity one second before it reaches the maximum height, (c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s. 

Answer:
(a) As in question number 23, maximum height
h = u²/2g. Here u = 28 m/s, take g = 9.8 m/s²
we get h = 28²/19.6 = 40 m

(b) Let t be the time to reach the maximum height when the final velocity is v = 0 m/s. From the equation v = u – gt,
we have  0 = u – gt
t = u/g

Now the time taken one second before maximum height = u/g – 1.

To get velocity at this time we again use the equation v = u – gt,

Now v = u – g(u/g – 1) = u – u + g = g = 9.8 m/s

(c) No, because the velocity in (b) is independent of u.

Q#26
A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.  

Answer:
Height of fall of balls are given by the equation,

h = ut + ½gt² = ½gt²   (Since u = 0)

For the 3rd ball t = 3 s, So h = ½(9.8)(9) = 44.1 m below the top of the building.

For the 4th ball t = 2 s, So h = ½(9.8)(4) = 19.6 m below the top of the building. 

For the 5th ball t = 1 s, So h = ½(9.8)(1) = 4.9 m below the top of the building. 

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