Q#1
A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field. (a) What distance he has to walk to reach the field? (b) What is his displacement from his house to the field?Answer:
(a) Total distance traveled by the man = 50 m + 40 m + 20 m = 110 m.
(b) His displacement is the vector joining points from his house to the field, which can be easily understood by the following figure:
Magnitude of displacement AB = √(30² + 40²) = 50 m
Angle with horizontal is given by tan θ = 30/40 = 3/4
Θ = tan-1(3/4) direction north to east.
Q#2
A particle starts from the origin, goes along the X-axis to the point (20 m, 0) and then returns along the same line to the point (-20 m, 0). Find the distance and displacement of the particle during the trip.
Answer:
Distance in forward motion = 20 m
Distance in backward motion = 20 m (up to origin) + 20 m (origin to further backward) = 40 m
Total Distance of the particle during the trip = 20 m + 40 m = 60 m
Displacement is the vector from initial position to final position, its magnitude = 20 m due negative direction of X-axis.
Q#3
It is 260 km from Patna to Ranchi by air and 320 km by road. An airplane takes 30 minutes to go from Patna to Ranchi whereas the deluxe bus takes 8 hours. (a) Find the average speed of the plane. (b) Find the average speed of the bus. (c) Find the average velocity of the plane. (d) Find the average velocity of the bus.
Answer:
(a) Average speed of the plane = 260 km/0.5 h = 520 km/h
(b) Average speed of the bus = 320 km/8 h = 40 km/h
(c) Average velocity of the plane is a vector with magnitude = Shortest distance divided by time taken = 260 km/0.5 h = 520 km/h and direction Patna to Ranchi straight.
(d) The average velocity of the bus is a vector with magnitude = Shortest distance divided by the time taken = 260 km/8 h = 32.5 km/h and direction Patna to Ranchi straight.
Q#4
When a person leaves his home for sightseeing in his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. (a) What is the average speed of the car during this period? (b) what is the average velocity?
Answer:
Distance covered by car = 12416 – 12352 = 64 km and time taken = 2 h
(a) Average speed of the car = 64 km/ 2 h = 32 km/h
(b) Displacement of the car is zero because it returns back to the same position, So average velocity = displacement/time = 0/2 h = 0
Q#5
An athlete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration?
Answer:
Difference of magnitudes of final and initial velocity = 18 – 0 = 18 km/h = 18000/3600 = 5 m/s, Time = 2 s
The magnitude of average acceleration =Difference of magnitudes of final and initial velocity divided by time = 5/2 = 2.5 m/s²
Q#6
The speed of a car as a function of time is shown in figure (3-E1). Find the distance traveled by car in 8 seconds and its acceleration.
Answer:
Distance traveled by car in 8 seconds is given by the area under the graph between 0 to 8 s. It is a triangle with base 8 s and height 20 m/s, Hence the area = ½ x 20 m/s x 8 s = 80 m
Acceleration of the car is given by the change of velocity divided by time interval = (20 – 0)/8 = 2.5 m/s².
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