Answer:
<h3>1.01 s</h3>
Explanation:
Using the equation of motion S = ut+1/2gt² to solve the problem where;
u is the initial velocity of the chocolate = 0m/s
t is the time taken
g is the acceleration due to gravity = 9.81m/s²
S is the height of fall = 5.0m
Substituting the given parameter into the formula to get the time t we have;
5 = 0(t)+1/2(9.81)t²
5 = 4.905t²
t² = 5/4.905
t² = 1.019
t = √1.019
t = 1.009 secs
<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>
Answer: it will move to the left
Explanation: i remeber doing this
The equation of the car is given by the equation,
x(t) = 2.31 + 4.90t² - 0.10t⁶
If we are going to differentiate the equation in terms of x, we get the value for velocity.
dx/dt = 9.8t - 0.6t⁵
Calculate for the value of t when dx/dt = 0.
dx/dt = 0 = (9.8 - 0.6t⁴)(t)
The values of t from the equation is approximately equal to 0 and 2.
If we substitute these values to the equation for displacement,
(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31
(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51
Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.