Topic 2.1: Displacement, Velocity and Acceleration Worksheet

IB PHYSICS GOHS

TOPIC 2.1: DISPLACEMENT, VELOCITY AND ACCELERATION WORKSHEET

1. A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 min at 80.0 km/h, 12.0 min at 100 km/h, and 45.0 min at 40.0 km/h, and spends 15.0 min eating lunch and buying gas. (a) Determine the average speed for the trip. (b) Determine the distance between the initial and final cities along this route. 2. (a) Sand dunes in a desert move over time as sand is swept up the windward side to settle in the lee side. Such “walking” dunes have been known to walk 20 feet in a year and can travel as much as 100 feet per year in particularly windy times. Calculate the average speed in each case in m/s. (b) Fingernails grow at the rate of drifting continents, about 10 mm/yr. Approximately how long did it take for North America to separate from Europe, a distance of about 3 000 mi? 3. A motorist drives north for 35.0 minutes at 85.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km in 2.00 h. (a) What is his total displacement? (b) What is his average velocity? 4. Figure P2.9 shows a position-time graph for several trains traveling along a straight track. Identify the graphs that correspond to (a) forward motion only, (b) backward motion only, (c) constant velocity, (d) greatest constant velocity, (e) no movement.

FIGURE P2.9 5. A tortoise can run with a speed of 0.10 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm). (a) How long does the race take? (b) What is the length of the race? 6. A particle is moving with a velocity of 60.0 m/s in the positive x direction at t = 0. Between t = 0 and t = 15.0 s the velocity decreases uniformly to zero. What was the acceleration during this 15.0-s interval? What is the significance of the sign of your answer?

7. A tennis ball with a speed of 10.0 m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 8.0 m/s. If the ball is in contact with the wall for 0.012 s, what is the average acceleration of the ball while it is in contact with the wall? 8. A certain car is capable of accelerating at a rate of +0.60 m/s2. How long does it take for this car to go from a speed of 55 mi/h to a speed of 60 mi/h? ANSWERS

1. Distances traveled are

x1 = v 1  t 1   80.0 km h 0.500 h  40.0 km

x2 = v 2  t 2   100 km h 0.200 h  20.0 km

x3 = v 3  t 3   40.0 km h 0.750 h  30.0 km

Thus, the total distance traveled is x 40.0  20.0  30.0 km  90.0 km , and the elapsed time is t 0.500 h  0.200 h  0.750 h  0.250 h  1.70 h .

x 90.0 km (a) v    52.9 km h t 1.70 h

(b) x  90.0 km (see above)

x 20 ft 1 m 1 yr  7 2. (a) v   7   2 10 m s , t 1 yr 3.281 ft  3.156  10 s 

or in particularly windy times

x 100 ft 1 m 1 yr  6 v   7   1 10 m s t 1 yr 3.281 ft  3.156  10 s 

(b) The time required must have been

3 3 x 3  10 mi 1609 m  10 mm  8 t      5 10 yr . v 10 mm yr 1 mi  1 m 

35.0  3. (a) Displacement 85.0 km h h   130 km 60.0  x 49.6  130 km  180 km

Displacement 49.6 130 km Average velocity    (b) elapsed time 35.0 15.0  63.4 km h  2.00  h 60.0 

4. (a) To correspond to only forward motion, the graph must always have a positive slope. Graphs of this type are a, e, and f . (b) For only backward motion, the slope of the graph must always be negative. Only graph c is of this type.

(c) The slope of the graph must be constant in order to correspond to constant velocity. Graphs like this are d and f .

(d) The graph corresponding to the largest constant velocity is the one with the largest constant slope. This is graph f .

(e) To correspond to no motion (i.e., zero velocity), the slope of the graph must have a constant value of zero. This is graph d .

5. (a) At the end of the race, the tortoise has been moving for time t and the hare for a time

t2.0 min  t  120 s . The speed of the tortoise is vt  0.100 m s , and the speed of the

hare is vh20 v t  2.0 m s . The tortoise travels distance xt , which is 0.20 m larger than

the distance xh traveled by the hare. Hence, xt x h  0.20 m , which becomes

vt t v h  t 120 s  0.20 m or

0.100 m st 2.0 m s t  120 s  0.20 m .

This gives the time of the race as t  1.3 102 s .

2 13 m (b) xt v t t 0.100 m s 1.3  10 s 

vf v i 0 60.0 m s 6. a     4.00 m s 2 tf t i 15.0 s  0

The negative sign in the above result indicates that the acceleration is in the negative x direction .

v v f i 8.0 m s   10.0 m s 3 2 7. a  2  1.5  10 m s tf t i 1.2  10 s

v v 60 55 mi h 0.447 m s  8. From a  , we have t    3.7 s t a 0.60 m s2  1 mi h 