1- A runner travels 10 km is 45 minutes. What is the runner\'s average speed in
ID: 1870231 • Letter: 1
Question
1- A runner travels 10 km is 45 minutes. What is the runner's average speed in km/hr?
2- There are two cars, Car A’s position can be modeled as xA = 2t2 + 3t + 2 and Car B’s position is xB = 35 t . Identify for each, when and if: there’s a turn around point, when they meet, who has acceleration.
3- A tiger jumps to catch a crane, if the tiger jumps at 30 degree angle respect to the ground at 25 m/s, what is the tiger’s highest point off the ground?
4- A motorbike is cruising at 20 m/s when suddenly the driver turns the engine up to full throttle. After traveling 5 km, the bike is moving with a speed of 60 m/s. What is the bike’s acceleration in m/s2 , assuming it is a constant acceleration?
5- An 6 hour trip is made at an average speed of 84 km/hr. If the driver drove at at 40 km/hr for the first 2.9 hours, what was the average speed (in km/hr) for the rest of the journey?
6- If you crash your car and it took you 2.5s to go from 60mph to stop, what is your acceleration in this case?
7- A ball is thrown upward, from a roof 90 m above the ground. The ball rises, then falls and strikes the ground some time later. The initial upward velocity of the ball is 53 m/s. Neglecting air resistance, after how many seconds will the ball strike the ground?
8- Joe drives due west at 35 km/hr for 70 minutes, then due north at 60 km/hr for 45 minutes then east at 55 km/hr for 4 hours. What is his displacement from his starting point?
Explanation / Answer
1. v = distance / time
v = 10 km / (45/60)hr
v = 13.33 km/h
2. vA = -4 t + 3
if it turn around then v= 0
vA = 0 at t = 3/4 sec
vB = d(xB)/dt = 35 , velocity is constant so it does not turn around.
when meet, xA = xB
-2 t^2 + 3 t + 2 = 35 t
2 t^2 + 32 t -2 = 0
t = 0.06 sec
a = dv/dt
so aA = -4 m/s^2
aB = 0
so car A has acceleration.
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