Question

1. If Mike drove for nine hours at a speed of 105km/h, how far did he drive in meters (m)?

2. Ann jogged 5.5 miles south to her house in 0.9 hours. What is Ann's speed? What is Ann's velocity?


3. How long will it take a car to accelerate from 18.9 m/s to33.7 m/s if the car has an acceleration of 4.2m/s2?

Answer 1

1) 945.000 meters

2) 9,84 kilometers/hour

3) 3,52 seconds

Step-by-step explanation:

1) 105 km/h x 9 h = 945km

1km = 1000 meters 945 km = 945.000 meters

2) 1 mile = 1.61 km

5.5 miles x 1.61 km / 1 mile = 8,855 km

You have to do a single third rule to clear the x which is the Ann's velocity in 1 hour.

8,855km in 0.9 hs

x (velocity in 1 hour) = 9,84 km = 1 hs

1 hs x 8,855km / 0,9hs = 9,84km/h

v= 9,84km/h

3) you have to use this formula and clear T (time)

vf = v0 + a.t

vf-v0 = a.t

(vf - v0)/a = t

(33.7 m/s - 18.9 m/s) / 4.2 m/s2 = 3,52 seconds

Answer 2

1) 945000 m

2) 6.11 miles/hour

3) 3.52 seconds

Step-by-step explanation:

1. Speed is the time rate of change of distance, it is the ratio of distance to time. Speed is given by the equation:

`speed(S)=(distance(d))/(time(t))`

Given that:

Time (t) = 9 hours and Speed (S) = 105 km/h

`S=(d)/(t)\\ d=S*t=105*9=945km=945000m`

2) Time (t) = 0.9 hours and distance (d) = 5.5 miles

`S=(d)/(t)\\ S=(5.5)/(0.9) =6.11miles/hour`

3) Acceleration it is the ratio of velocity to time. Acceleration is given by the equation:

`Accceleration(a)=(Final,velocity(u)-initial, velocity(v))/(time(t))`

Given that :

a = 4.2 m/s² , u = 18.9 m/s and v = 33.7 m/s

`a=(v-u)/(t)\\ t=(v-u)/(a)\\ t=(33.7-18.9)/(4.2)=3.52s`