Question

A car moving with a velocity of 10 m/s accelerates uniformly at 1m/s2 until it reaches a velocity of 15 m/s. Calculate (i) the time taken (ii) the distance traveled during the acceleration (iii) the velocity reached 100m from the place where the acceleration began.

Answer 1

The movement with uniform acceleration is described by expressions:

$v=v_0+at$, (1)

$s= v_0t+\frac{at^2}{2}$ , (2)

where $v$ - velocity;

$v_0$ - initial velocity;

$a$ - acceleration;

$s$ - distance;

$t$ - time.

1. The time taken for increasing velosity from 10 to 15 m/s can be found using (1):

$t=\frac{v-v_0}{a}=\frac{15-10}{1}= 5\space s$.

2. The distance traveled during the acceleration can be found using (2):

$s= v_0t+\frac{at^2}{2}=10 \cdot 5+\frac{1 \cdot 5^2}{2}=62.5 \space m.$

3. The velocity reached 100 m from the place where the acceleration began can be found in such way/

First, find the time, required for moving on 100 m, from (2):

$100= 10t+\frac{t^2}{2},$ or in form of the reduced quadratic equation:

$t^2+20t-200=0.$

The roots of this equation $t=\frac{1}{2}(-20 \pm \sqrt (20^2-4 \cdot (-200)))$ .

$t_1=7.32 \space s,$

$t_2=-27.32 \space s$ (has no physical meaning).

Then, using (1), find reached velosity:

$v=v_0+at = 10+1 \cdot 7.32=17.32 \space m/s.$

Answer:

(i) 5 s,

(ii) 62.5 m,

(iii) 17.32 m/s.