Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:
<u>v₀ = 16.55 m/s</u>
Answer:
a) 
b) 
c) 
d) Displacement = 22 m
e) Average speed = 11 m/s
Explanation:
a)
Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:
Therefore, acceleration is
b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:
c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:
d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):
Displacement = 
e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:
Average velocity = 
