laws of motion calculation

Question

Not sure about the answer to this question, it seems to indicate an illogical length.

Using Laws of Motion a) A jet aeroplane has a landing speed of $70m/s$. Its speed is reduced to $8m/s$ in $10$ seconds. Calculate the distance traveled in this time.

b) When a jet is taking off, from a stationary point, it must reach $80m/s$. It has to accelerate at $1m/s^2$. Calculate the minimum length of runway.

a)

$v=u+at$

$v=8m/s$

$u=70m/s$

$t=10s$

Find $a$

$8= 70 + a10$

$8 - 70 = a10$

$(8-70)/10 = a$

$a = -6.2m/s2$

$d=ut + 0.5 at^2$

$u=70m/s$

$v=8m/s$

$t=10s$

$a= -6.2m/s$

Find $d$

$d = (70)(10) + (0.5)(-6.2)(10)2$

$d = (700) + (-310)$

$d = 390m$

Aircraft travels $390m$

b)

$v2 = u2 + 2ad$

$u=0m/s$

$v=80m/s$

$a=1m/s$ Find $d$

$(80)2 = (0)2 + (2)(1)d$

$(80)2 - (0)2/ (2)(1) = d$

$6400/2 = d$

$d= 3200$

Runway is $3200m$ (Seems wrong)

Answer 1

Both of your answers in fact are correct (Since we assume uniform acceleration).

---

For (a), your answer is fine. Here is another one of Newton's laws of constant acceleration (A big shortcut to your method): $$s=\frac{1}{2}(u+v)t$$ Which gives: $$s=\frac{1}{2}(70+8)\cdot 10=39\cdot 10=390 \text{ m}$$

---

For (b), you used $v^2=u^2+2as$ which works fine.

You can also obtain the same answer by using $s=ut+\frac{1}{2}at^2$. For a $a=1 \text{ ms}^{-2}$ acceleration, it will take $t=80 \text{ s}$ to reach $v=80 \text{ ms}^{-1}$.

Therefore: $$s=0+\frac{1}{2}\cdot 1\cdot 80^2=\frac{6400}{2}=3200 \text{ m}$$ It would be interesting to know what the answer key suggests (If you have any).

Answer 2

a) $v-u=at$

$a=\frac{v-u}{t}$

$a=\frac{8-70}{10}=-6.2m/s^2$

$d=ut+\frac{1}{2}at^2=t(u+\frac{1}{2}at)=10(70-31)=390m$

b)

$t=v/a=\frac{80}{1}=80s$

For uniform acceleration, average speed is useful: On reaching $80m/s^2$ from standstill, average speed will have been $40m/s^2$

$40m/s\times 80s=3,200m$