Question

A car starting motion which has an initial velocity 5m/s and accelerate at a rate of 10m/s square. Calculate the a) The time after a distance of 100m.
b) The final velocity after 100m.

Answer 1

Question :

A car starting motion which has an initial velocity 5m/s and accelerate at a rate of 10m/s square.

To Find :

a) The time after a distance of 100m.

b) The final velocity after 100m.

Formula Used :

There are three equations of motion which are derived from Newton's first law of motion :-

$v = u + at$

$s = ut + \frac{1}{2} a {t}^{2}$

$2as = {v}^{2} - {u}^{2}$

Here, a is acceleration, u is initial velocity, v is final velocity and s is the distance.

Solution :

(a) The time after a distance of 100m.

By using second equation of motion,

$s = ut + \frac{1}{2} a {t}^{2} \\ 100 = 5t + \frac{1}{2} \times 10 {t}^{2} \\ 100 = 5t + 5 {t}^{2} \\ 20 = t + {t}^{2}$

Now it becomes in the form of a quadratic equation,

${t}^{2} + t - 20 = 0 \\ {t}^{2} + 5t - 4t - 20 = 0 \\ t(t + 5) - 4(t + 5) = 0 \\ (t + 5)(t - 4) = 0$

So, t = - 5 & t = 4. As time cannot be negative then t = 4 seconds.

(b) The final velocity (v) after 100m.

By using third equation of motion,

$2as = {v}^{2} - {u}^{2}$

$2 \times 10 \times 100 = {v}^{2} - {5}^{2} \\ 2000 = {v}^{2} - 25 \\ 2025 = {v}^{2} \\ \sqrt{2025} = v \\ 45 \: m {s}^{ - 1} = v$