A car starts from rest and accelerates uniformly at 2.5 m/s^2 for 20.0 s, travel

Question

A car starts from rest and accelerates uniformly at 2.5 m/s^2 for 20.0 s, travels with constant speed for 10.0 minutes, and finally decelerates at 5.0 m/s^2 until it stops. (a) What is the average speed of the car during the entire period? (b) How much total distance does it travel from the start to the end? Starling from rest, a motorcycle driver accelerates uniformly at 4.0 m/s^2 for 15.0 s and then slows down with an acceleration of-4.0 m/s^2 for the next 15.0 s. (a) What is the maximum speed of the motor cycle attained during the trip? (b) How far does he travel during the entire trip? During the Fourth of July fireworks show, a firecracker launched at an angle of 60.0 degree above the horizontal at a speed of 48.0 m/s explodes in the air 6.0 s after it is launched, (a) How high was it above the ground when it exploded? (b) What was its velocity (magnitude and direction) at the time it exploded? Students from a 20.0 m high second floor window of a fraternity house use a hose to throw water on the students in the other fraternity 40.0 m away across the street. Water comes out of the hose with a speed 20.0 m/s at an angle of 30.0 degree with the horizontal, (a) How high above the ground will the water strike the wall of the other fraternity? (b) At what angle will the water stream strike the wall? A hunter aims his rifle onto his target at a distance of 200 m running away from the hunter at a speed of 5.0 m/s. He aims his rifle at an angle of 45.0 degree with respect to the horizontal, (a) What should be the speed of the bullet fired from the rifle if it has to hit the running target? (b) What is the distance of the target, from the hunter, when it is struck by the bullet? (Neglect the height of the hunter and air friction.)

Explanation / Answer

10. When accelerating,

d = u t + a t^2 /2

d1 = 0 + (2.5 x 20^2) /2 = 500 m

velocity after 20s, v = u + a t = 0 + 2.5 x 20 = 50 m/s

For constant speed period,

d2 = (50 m/s ) (10 x 60 s) = 30000 m

When decelrating,

v = u + a t

0 = 50 - 5t

t = 10sec

d3 = (50 x 10) - (5 x 10^2 / 2) = 250 m

(A) average speed = total distance travelled / time

= (500 + 30000 + 250) / (20 + 600 + 10)

= 48.8 m/s

(B) total distance travelled= d1 + d2 +d3 = 30750 m

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