Please show every step in the solutions for the problems! Many thanks in advance

Question

Please show every step in the solutions for the problems! Many thanks in advance!

If the average velocity of an object is zero in some time interval, what can you say about its displacement in that interval? What about its distance travelled? Explain your answers. Is it possible for a particle to have a velocity and an acceleration in opposite directions? If so, sketch a velocity-time graph proving your point. If not, explain why not. A runner runs in a straight line with an average velocity of or 5 minutes and then with an average velocity of for 4 minutes, (a) What is her total displacement? (b) What is her average velocity during this time, (c) Suppose her goal is to run 5 km in 15 minutes. What average velocity must she have over the remaining portion of her run? A ball is thrown upward. While the ball is in the air, (a) does its acceleration increase, decrease or remain the same? Explain your answer, (b) Describe what happens to its velocity. The position of a pine wood derby car was observed at various times as shown in the table below. Find the average velocity of the car for (a) the first second, (b) the last 3 seconds and (c) the entire period of observation.

Explanation / Answer

1. Since the average velocity is zero that means the total displacement is zero for the object because we know average velocity = total displacement/total time taken

Displacement is the direct distance between initial and final points but for distance covered we take into account the length of the path taken from initial to final point so in case of displacement as zero it is quite possible that distance travelled is not zero.for example if you make a roun trip your displacement would be zero but not distance.

2. yes velocity and acceleration can be opposite as acceleration is change in velocity and it is in the direction of net force on the body so if the force on body is acting opposite to its velocity then the acceleration and velocity will be in opposite direction which is the case when we throw a ball up in the air its initial velocity is upwards but the gravitational pull on the object is downwards and the acceleration of object is g downwards. So with time the velocity of object decreases, becomes zero at the highest point reached by the object and then object starts falling downward with its velocity increasing in downward direction.

3. (a) total displacement in x direction = 5(5x60)=1500 i

total displacement in y direction =4(4x60) =960 j

net displacement = 1500i+960j

magnitude of displacement =(1500^2+960^2)^(1/2) =1780.90 m

(b) average velocity during the time = 1780.90/(9x60)=3.3m/s

(c) remaining distance to be covered = 5000-1780.9 =3219.1 m

remaining time = 15-9 =6 min = 360 sec

average velocity required = 3219.1/360 =8.94m/s

4. (a) when the ball is in air the only force acting on it is gravitational force (weight) which is constant so the cceleration of the ball will be constant

(b) since initial velocity of ball is upwards and gravitational accelration is downwards so with time the velocity of ball will decrease and after some time it will become zero then the ball will start falling downwards .In the downward motion the gravitational acceleration will increase its velocity as it is now in the same direction as the velocity.

5.

(a) for first second avg velocity = (displacement till first second)/time = 2.3/1 = 2.3 m/s

(b) last 3 seconds aveg. velocity = (displacement in last 3 seconds)/3 =(57.5-9.2)/3 =16.1m/s

(c) for the entire observation avg. veloctiy = total displacement/total time = 57.5/5 =11.5 m/s

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