5.)A person takes a trip, driving with a constant speed of 98.5 km/h, except for

Question

5.)A person takes a trip, driving with a constant speed of 98.5 km/h, except for a 30.0-min rest stop. The person's average speed is 66.0 km/h.

(a) How much time is spent on the trip?
h =

(b) How far does the person travel?
km=

6.)A 53.5-g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 18.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.90 ms, what is the magnitude of the average acceleration of the ball during this time interval?
m/s2=

7.)A particle starts from rest and accelerates as shown in the figure below.

(a) Determine the particle's speed at t = 10.0 s.
m/s=

Determine the particle's speed at t = 20.0 s?
m/s=

(b) Determine the distance traveled in the first 20.0 s. (Enter your answer to one decimal places.)
m=

8.)A particle moves along the x axis according to the equation

x = 1.99 + 2.93t 1.00t2,

where x is in meters and t is in seconds.

(a) Find the position of the particle at t = 2.60 s.
m=

(b) Find its velocity at t = 2.60 s.
m/s=

(c) Find its acceleration at t = 2.60 s.
m/s2=

Explanation / Answer

(5) Let us assume that the person travel for "t" hr in the trip.
Then the distance covered will be = 98.5*t km
And the time for the trip will be = t+ (30/60) = (t+0.5) hr
Hence we know that the average speed = total distance /total time
Average speed = 66 km/h , therefore
66 = 98.5t /(t+0.5)
t+0.5 = 1.49t
0.49t = 0.5
t = 1.015 hr
Hence he spent total time = t + 0.5 = 1.015+0.5 = 1.515 hr
(B) Distance covered will be = 98.5*t = 98.5*(1.015) = 100.015 km

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