Question

a car moving with a velocity pf 36 km/h-1 is brought to rest in s5 seconds,calculate its decleration?

Answer 1

A car moving with a velocity of $36~\mathrm{km/h}$ is brought to rest in s5 seconds. Calculate its deceleration?

**Solution.**

$$v_i = 36 \frac{\mathrm{km}}{\mathrm{h}}, \quad t = 5s;$$ $$a - ?$$

A velocity of the car:

$$v = v_i - a t.$$

$v_i$ – the initial velocity;

$v$ – the final velocity;

$a$ – the deceleration;

$t$ – the time.

Converting the initial velocity to meters per hour:

$$v_i = 36 \frac{\mathrm{km}}{\mathrm{hour}} \left( \frac{1000m}{1\mathrm{km}} \right) = 36000 \frac{\mathrm{m}}{\mathrm{hour}}.$$

Converting the initial velocity to meters per second:

$$v_i = 36000 \frac{\mathrm{km}}{\mathrm{hour}} \left( \frac{1\mathrm{hour}}{3600\mathrm{second}} \right) = 10 \frac{\mathrm{m}}{\mathrm{second}}.$$ $$v_i = 10 \frac{\mathrm{m}}{\mathrm{s}}.$$

A car is brought to rest then:

$$v = 0.$$ $$0 = v_i - a t;$$ $$v_i = a t;$$ $$a = \frac{v_i}{t}.$$

A deceleration is:

$$a = \frac{10 \frac{\mathrm{m}}{\mathrm{s}}}{5s} = 2 \frac{\mathrm{m}}{\mathrm{s}^2}.$$

**Answer:** A deceleration is $a = 2\frac{\mathrm{m}}{\mathrm{s}^2}$.