# A car's begin speed is 70 km/h and it accelerates with 6000 km/h^2 . How many seconds does it take to get from 70 km/h to 120 km/h and what distance did the car need ?

Sep 27, 2017

The speed increase required 72 seconds and 1.9 km.

#### Explanation:

We can determine the time using the suvat formula
$v = u + a \cdot t$

Plugging in the known data and solving for t

$120 \frac{k m}{h} = 70 \frac{k m}{h} + 6000 \frac{k m}{h} ^ 2 \cdot t$

$t = \frac{120 \frac{k m}{h} - 70 \frac{k m}{h}}{6000 \frac{k m}{h} ^ 2}$

t = (50 (cancel(km))/cancel(h) ) / (6000 (cancel(km))/(h*cancel(h)

$t = 0.00833 h$

Converting to seconds: $0.00833 h \cdot \frac{3600 s}{1 h} = 30 s$

We can determine the distance using the suvat formula
$s = \left(\frac{u + v}{2}\right) \cdot t$
Notice that this formula calculates the average velocity and multiplies by time. This is legitimate only because the acceleration is constant throughout.

Plugging in the known data
$s = \left(\frac{70 \frac{k m}{h} + 120 \frac{k m}{h}}{2}\right) \cdot 0.00833 h$

$s = 95 \frac{k m}{h} \cdot 0.00833 h = 0.79 k m$

I hope this helps,
Steve