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Chapter 2 Question 5: The position of an object moving along an x axis is given...


Chapter 2
Question 5: The position of an object moving along an x axis is given by, where x is in meters and t is in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement between and s? (f) What is its average velocity for the time interval from s to s?
Question 6: A pickup vehicle is moving with a speed of 15.00 m/s on a straight road. A scooterist wishes to overtake the pickup vehicle in 150.0s. If the pickup vehicle is at an initial distance of 1.500 km from the scooterist, with what constant speed should the scooterist chase the pickup vehicle?
Question 7: A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?
Question 8: Raindrops fall 1800 m from a cloud to the ground. (a) If they were not slowed by air resistance, how fast would the drops be moving when they struck the ground? (b) Would it be safe to walk outside during a rainstorm?
Question 9: Rachel walks on a straight road from her home to a gymnasium 2.80 km away with a speed of 6.00 km/h. As soon as she reaches the gymnasium, she immediately turns and walks back home with a speed of 7.70 km/h, as she finds the gymnasium closed. What are the (a) magnitude of average velocity and (b) average speed of Rachel over the interval 0.00-35.0 min?
Question 10: A car m
oves uphill at 35 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
Question 11: The displacement of a particle moving along an x-axis is given by , where x is in meters and t is in seconds. Calculate (a) the instantaneous velocity at and (b) the average velocity between and .

Solutions

Expert Solution

Question-5

We know that

where

S=distace

U= Initial velocity

a=acceleration

t=time

a)

for t=1s

S=U+0.5a m

if acceleration is equal to zero then

S=U

b)

for t=2s

S=2U+0.5*a*^=2U+2a

if acceleration is equal to zero then

S=2U

c)

for t=3s

S=3U+0.5*a*3^2=3U+4.5a

if acceleration is equal to zero then

S=3U

e)

the displacement between time t1 and t2 is given by

displacement=S1-S2=(Ut1+0.5*a*t12)-(Ut2+0.5*a*t22)

if acceleration is equal to zero then

displacement=U(t1-t2) m

f)

the average velocity

v=(s1t1+s2t2)/(t1+t2)

where ' s1 ' and ' s2 ' are the displacement of the object at time ' t1 ' and ' t2 ' respectively

Question-2

the velocity of the pickup vehicle is 15.0m/s=V (say)

at t=0 the distance between the pickup vehicle and scooter is 1.5km=1500m=d (say)

at time t=150.0s, they will meet if the scooter moves with ' U ' velocity

so, we can write

Ut=Vt+1500

or,

U*150=15.0*150+1500

or,

U=15.0+10

or,

U=25.0m/s


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