Question

An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates from 54 m/s to a complete stop. How much distance did it travel during that acceleration?

Answer 1

Answer:

Displacement (s) of an object equals, velocity (u) times time (t), plus ½ times acceleration (a) times time squared (t2). Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

Answer 2

Given:

• Acceleration = a = -6.8 m/
• ${s}^{2}$
• Initial velocity = u= 54m/s
• Final velocity = v = 0m/s

━━━━━━━━━━━━━━━

Need to find:

• The distance travelled= ?

━━━━━━━━━━━━━━━

PROCESS I:

• Find the time taken to stop first
• And then calculate the distance travelled

From 1st equation of motion we know,

$v = u + at$

$\implies \: 0 = 54 + ( - 6.8)t$

$\implies \: t = 7. 9sec$

We know from 2nd equation of motion,

$s = ut + \dfrac{1}{2} a {t}^{2}$

$\implies \: s = 54 \times 7.9 + 0.5 \times ( - 6.8) \times {7.9}^{2}$

$\implies \: s = 426.6 - 212.194$

$\red{\boxed{\large{\bold{ \implies \: s \approx \: 214.4 \: m}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

PROCESS II:

Use 3rd equation of motion:

${v}^{2} - {u}^{2} = 2as$

$\implies {0 }^{2} - {54}^{2} = 2 \times ( - 6.8) \times s$

$\implies \: s = \dfrac{2916}{2 \times 6.8}$

$\red{\boxed{\bold{\large{ \implies \: s \approx \: 214.4m}}}}$