# If acceleration is increasing at the constant rate of 2 m/s^2. Find the distance travelled in 5 sec? If initial velocity and acceleration both were zero. (A)  75m (B) 100m  (C) 125/3 m

Mar 22, 2017

After 5 seconds the object will have reached a speed of:
$v = 5 s \times 2 m / {s}^{2} = 10 m / s$

#### Explanation:

Since the speed goes up from 0 to 10 m/s at a constant rate, the average speed in those 5 seconds will be:

$\overline{v} = \frac{0 + 10}{2} m / s = 5 m / s$

Since distance=(average speed) times (time):

Distance $s = \overline{v} \times t = 5 \frac{m}{\cancel{s}} \times 5 \cancel{s} = 25 m$

I'm afraid that's not one of the choices you had... but I'm pretty sure it's correct.

Mar 24, 2017

$\Rightarrow x = \frac{125}{3}$

#### Explanation:

$\frac{\mathrm{da}}{\mathrm{dt}} = 2$

$\Rightarrow \mathrm{da} = 2 \mathrm{dt}$

$\Rightarrow {\int}_{0}^{a} \mathrm{da} = 2 {\int}_{0}^{t} \mathrm{dt}$

$\Rightarrow a = 2 t$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}} = 2 t$

$\Rightarrow \left(\mathrm{dv}\right) = 2 t \mathrm{dt}$

$\Rightarrow {\int}_{0}^{v} \left(\mathrm{dv}\right) = 2 {\int}_{0}^{t} \cdot \mathrm{dt}$

$\Rightarrow v = 2 {t}^{2} / 2$

$\Rightarrow v = {t}^{2}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} = {t}^{2}$

$\Rightarrow \mathrm{dx} = {t}^{2} \cdot \mathrm{dt}$

$\Rightarrow {\int}_{0}^{x} \mathrm{dx} = {\int}_{0}^{5} {t}^{2} \cdot \mathrm{dt}$

$\Rightarrow x = {5}^{3} / 3$

$\Rightarrow x = \frac{125}{3}$