4- A certain corner of a room is selected as the origin of a rectangular coordin

Question

4- A certain corner of a room is selected as the origin of a rectangular coordinate system. A fly is crawling on an adjacent wall at a point having coordinates (1.9, 1.3), where the units are meters. Express the location of the fly in polar coordinates. (a) Find the truck's original speed.
m/s

(b) Find its acceleration.
m/s2
6- A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 120 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is? = 32.3°. How wide is the river? 7- certain car is capable of accelerating at a rate of 0.59 m/s2. How long does it take for this car to go from a speed of 56 mi/h to a speed of 65 mi/h? r = m ? = °

5- A truck covers 40.0 m in 9.20 s while uniformly slowing down to a final velocity of 1.60 m/s. (a) Find the truck's original speed.
m/s

(b) Find its acceleration.
m/s2
6- A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 120 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is? = 32.3°. How wide is the river? y = m 7- certain car is capable of accelerating at a rate of 0.59 m/s2. How long does it take for this car to go from a speed of 56 mi/h to a speed of 65 mi/h? s A certain corner of a room is selected as the origin of a rectangular coordinate system. A fly is crawling on an adjacent wall at a point having coordinates (1.9, 1.3), where the units are meters. Express the location of the fly in polar coordinates.

Explanation / Answer

1)
The acceleration of the object must remain constant <--- only this is tue


2)
It eventually stops and then speeds up in the southward direction

3)

a)KE = P^2 / ( 2m )

so.. using dimensional analysis..

kg. * m^2 / s^2 = P^2 / kg

so.. P^2 = kg^2 * m^2 / sec^2
so.. dimesnion of P = kg*m/sec .... option B


4)

r = sqrt ( 1.9^2 + 1.3^2 ) = 2.302173 m

theta = tan inverse ( 1.3 / 1.9 ) = 34.3803447 degree


5)

a)
Let the acceletioan be a .. and inital veloctu be u ..
final velocity = 1.6 m/sec
distance s = 40 m
time t = 9.2 secs

s = ut + 0.5 * a*t^2
so.. 40 = 9.2 u + 0.5*a * 9.2^2
so.. a = 0.945179584 - 0.21739130435 * u

also..
v = u + at
1.6 = u + (0.945179584 - 0.21739130435 * u )*9.2

so.. inital speed , u = 7.0956521728 m/sec

b) acceleraion a = 0.945179584 - 0.21739130435 * u = -0.597353497 m/sec2



6)

as shown in the figure ..
x = 120 m
theeta = 32.3 degree

so.. tan theeta = y / x

so. tan 32.3 = y / 120
so.. y= 75.86085093 m

so.. widhth of river = 75.86085093 m

7)
56 mi / hr = 25.0342 m/sec
65 mi/hr = 29.0576 m/sec

so.. initial speed = u = 25.0342 m/sec
final speed = v = 29.0576 m/sec
acceleraion = 0.59 m/sec2

let the time be t ..
so. t = ( v - u ) / a = ( 29.0576 - 25.0342 ) / 0.59 = 6.819322 secs

so.. time = 6.819322 secs

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