A hiker walks with an average speed of 1.8 m/s. What distance in kilometers does

Question

A hiker walks with an average speed of 1.8 m/s. What distance in kilometers does the hikertravel in a time of 88 min?
1 km
A car traveling with an initial velocity of 16 m/s accelerates at a constant rate of 4 m/s2 for a time of 4 seconds. (a) What is its velocity at the end of thistime?
1 m/s

(b) What distance does the car travel during this process?
2 m A car moving with an initial velocity of 30 m/s slows down at a constant rate of -3 m/s2. (a) What is its velocity after 3 seconds ofdeceleration?
1 m/s

(b) What distance does the car cover in this time?
2 m


A hiker walks with an average speed of 1.8 m/s. What distance in kilometers does the hikertravel in a time of 88 min?
1 km
A car traveling with an initial velocity of 16 m/s accelerates at a constant rate of 4 m/s2 for a time of 4 seconds. (a) What is its velocity at the end of thistime?
1 m/s

(b) What distance does the car travel during this process?
2 m A car moving with an initial velocity of 30 m/s slows down at a constant rate of -3 m/s2. (a) What is its velocity after 3 seconds ofdeceleration?
1 m/s

(b) What distance does the car cover in this time?
2 m

(a) What is its velocity at the end of thistime?
1 m/s

(b) What distance does the car travel during this process?
2 m (a) What is its velocity after 3 seconds ofdeceleration?
1 m/s

(b) What distance does the car cover in this time?
2 m

Explanation / Answer

Average speed (v) of the hiker = 1.8 m/s Time (t) he travelled = 88 min                                = (88)(60s) = 5280 s Now the distance (d) he traveled in the given time is                       d = vt                          = (1.8m/s)(5280s)                          = 9504 m                          = 9.504*103m                          = 9.504 km Please ask remaining questions in the Physics Q & A Board asseparate.

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