Calculate the time taken for a car to cover a distance of 125m if the initial speed is 5m/s and it has constant acceleration of 1.5m/s2?
d = 1/2 a t^2 + Vo t
125 = .75 t^2 + 5 t ... solve for t ... quadratic formula probably best method
To calculate the time taken for a car to cover a distance, we can use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Where:
s = distance covered
u = initial speed (velocity)
t = time taken
a = acceleration
Given values:
Initial speed, u = 5 m/s
Acceleration, a = 1.5 m/s²
Distance, s = 125 m
To find the time taken, we rearrange the equation and solve for t:
\(s = ut + \frac{1}{2}at^2\)
Substituting the given values:
125 = 5t + (1/2)(1.5)t^2
Now we have a quadratic equation in terms of t. Rearrange the equation to bring it in standard form:
0 = (1/2)(1.5)t^2 + 5t - 125
To solve this quadratic equation, we can either factorize it or use the quadratic formula. In this case, let's use the quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For the equation 0 = (1/2)(1.5)t^2 + 5t - 125,
a = (1/2)(1.5) = 0.75
b = 5
c = -125
Substituting these values into the quadratic formula, we get:
\(t = \frac{-5 \pm \sqrt{5^2 - 4(0.75)(-125)}}{2(0.75)}\)
\(t = \frac{-5 \pm \sqrt{25 + 375}}{1.5}\)
\(t = \frac{-5 \pm \sqrt{400}}{1.5}\)
Simplifying further:
\(t = \frac{-5 \pm 20}{1.5}\)
Now, we have two possibilities, one with a plus sign and one with a minus sign:
\(t_1 = \frac{-5 + 20}{1.5} = \frac{15}{1.5} = 10\) seconds
\(t_2 = \frac{-5 - 20}{1.5} = \frac{-25}{1.5} \approx -16.67\) seconds
Since time cannot be negative in this context, we discard the negative value. Therefore, the time taken for the car to cover a distance of 125m is approximately 10 seconds.