Question

1) A car is traveling at +15m/s when a constant acceleration of -4m/s^2 (say a strong continuous wind opposite the initial direction of motion) is applied. (Note: No friction and the car is coasting before the wind starts.) 5 seconds later, what are the position, velocity, and acceleration of the car?

2) A 0.44kg size 5 soccer ball is kicked from the top of a 20m cliff with a velocity of 10 m/s at an angle of 35° with respect to the ground. What is the maximum height the ball reaches (above the ground) and what is the horizontal distance it travels when it eventually hits the ground?

3) If you walk to a store that is 2 miles away in 1 hour, and then walk home from the store in only 1/2 of an hour, compute the average velocity for the entire trip.

Answer 1

1. u=15 m/s t=5s
   a= -4 m/s²
v=u+at
v=15-4*5
v= -5 m/s (final velocity)
s=ut+(0.5)*(at²)
s=15*5 + (0.5)*(-4*5*5)
s= 25 m (position)
since accelaration is constant, hence after 5 s accelaration is -4 m/s2

2.) horizontal velocity from top of cliff (u_x)= 10*cos(35) = 8.1912 m/s
   vertical velocity from top of cliff (u_y) = 10*sin(35) = 5.7357 m/s
maximum height it reaches is when vertical velocity becomes 0
h_max= height of cliff + (u²_y)/2g
= 20 + (5.7357^2)/2*10
   = 21.6449 m
to reach ground, displacement is -20 m
s=u_yt+(0.5)*(at²)
-20=5.7357*t + (0.5)*(-10 t²)
5t² - 5.7357t -20 =0
solving we get t=2.6541 s
horizontal dist = u_x * t = 8.1912 * 2.6541 =21.74 m

3.) since we start from home and end at home hence net displacement is 0.
avg velocity = (avg displacement)/(time take)
avg velocity = 0 m/s