Name: Solving Free Fall Problem Using Quadratic Equation
Uploaded: 2019-07-19
Duration: 7 min 7 s
Description: In this video, we will learn how we can solve the problem of free fall effectively using the quadratic equation. Let's watch the video and learn.

Question

A cyclist is travelling at a velocity of

10 m s^{- 1}

when he reaches the top of a slope, which is

80 m

long. There is a bend at the bottom of the slope, which it would be dangerous to go round any faster than

11 m s^{- 1} .

Because of gravity, if he did not pedal or brake he would accelerate down the slope at

0.1 m s^{- 2} .

To go as fast as possible but still reach the bottom at a safe speed should the cyclist brake, do nothing or pedal?

Accepted Answer

Given,

Initial velocity, $u = 10 m s^{- 1}$

Length of slope, $s = 80 m$

Acceleration down the slope without pedalling, $a = 0.1 m s^{- 2}$

For uniform acceleration,

Acceleration is given by, $a = \frac{v - u}{t}$ $— equation (1)$

Substitute the values,

$\Rightarrow 0.1 = \frac{v - 10}{t}$

$\Rightarrow v = 0.1 t + 10$ $— equation (2)$

Now, from suvat equations,

Displacement, $s = \frac{1}{2} (u + v) t$ $— equation (3)$

Substitute the values and also put the value of $v$ from $equation (2)$ .

$\Rightarrow 80 = \frac{1}{2} (10 + 0.1 t + 10) t$

$\Rightarrow 160 = 20 t + 0.1 t^{2}$

$\Rightarrow 0.1 t^{2} + 20 t - 160 = 0$

Using quadratic formula, we get

$\Rightarrow t = \frac{- 20 \pm \sqrt{400 - 64}}{0.2}$

$\Rightarrow t = \frac{- 20 \pm 21.54}{0.2} = 7.7 s$ and $- 207.7 s$

$($ We are ignoring the negative value as time cannot be negative. $)$

$∴ t = 7.7 s — equation (4)$

Substituting the value of time from $equation (4)$ to $equation (2)$ , we get,

$v = 0.1 \times 7.7 + 10 = 10.77 m s^{- 1} \approx 10.8 m s^{- 1}$ .

If he is not pedalling then he will arrive at the bend with velocity $10.8 m s^{- 1}$ which is in safe limits as it is still less than $11 m s^{- 1}$ . So he can pedal.

Important Questions on Velocity and Acceleration

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