Using the Equation:
                                 v² = vi² + 2 · a · s    → Eq.1
where,
v = final velocity 
vi = initial velocity 
a = acceleration 
s = distance 
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
 
</span></span> v² = vi² + 2 · a · s
 v² = 0 + 2 · a · s
 v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s 
Substituting the values,
a = 46</span>²/2×1<span> 
a = 1058 m/s</span>² 
<span>Now applying Newton's 2nd law of motion,
 </span>
<span>F = ma
   = 0.145</span>×<span>1058
F = 153.4 N</span>
        
             
        
        
        
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s 
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =? 
Initial velocity (U) =?
A. Determination of the time taken for the car to stop. 
Let us obtain an express for time (t) 
Acceleration (a) = Velocity (V)/time(t) 
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t 
a = s/t^2
Cross multiply 
a x t^2 = s
Divide both side by a 
t^2 = s/a
Take the square root of both side 
t = √(s/a) 
Now we can obtain the time as follow 
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..? 
t = √(s/a) 
t = √(256/26) 
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car. 
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2 
Time (t) = 3.14 sec 
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms 
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
 
        
             
        
        
        
A :-) it was given the name Newton (N). from this, the derived unit of energy (or work) is defined ,as the work produced when the unit of force causes a displacement equal to the unit of length of its point of application along its direction . It was given the name Joule (J).