Question

(correct answers only)

Answer 1

• Initial velocity=11.8m/s=u
• Acceleration=a=6.39m/s²
• Distance=s=55m
• Time be t

According to second equation of kinematics

• s=ut+1/2at²
• 55=11.8t+1/2(6.39)t²
• 110=23.6t+6.39t²
• 6.39t²+23.6t-110=0

Take positive value

• t=2.7s

Answer 2

Answer:

2.7 s  (1 d.p.)

Step-by-step explanation:

Use the Constant Acceleration Equations where:

• s = displacement in meters
• u = initial speed in m/s
• v = final speed in m/s
• a = acceleration in m/s²
• t = time in s (seconds)

Given:

• s = 55 m
• u = 11.8 m/s
• a = 6.39 m/s²

$\begin{aligned}\text{Using }s & =ut+\dfrac{1}{2}at^2\\\implies 55 & =11.8t+\dfrac{1}{2}(6.39)t^2\end{aligned}$

Rearrange the equation to equal zero:

$\implies 3.195t^2+11.8t-55=0$

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Use the quadratic equation to solve for t:

$\implies a=3.195, \quad b=11.8, \quad c=-55$

$\implies t=\dfrac{-11.8 \pm \sqrt{11.8^2-4(3.195)(-55)} }{2(3.195)}$

$\implies t=2.694780675, -6.388051411$

As time is positive:

$\implies t=2.694780675... \:\text{s}$

$\implies t=2.7 \: \text{s (1 d.p.)}$

Learn more about SUVAT equations here:

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