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 car is at rest when it accelerates at

5 m s^{- 2}

for

4 s .

It then continues at a constant velocity. At the instant the car starts moving, a truck passes it, moving at a constant speed of

22 m s^{- 1} .

After

10 s

the truck starts slowing at

1 m s^{- 2}

until coming to rest.
Show that the distance covered at a time

t s

from the start by the car and the truck, for

t > 10,

are given by

40 + 20 (t - 4)

and

220 + 22 (t - 10) - \frac{1}{2} (t - 10)^{2},

respectively. Hence, find the time at which the car passes the truck.

Accepted Answer

Given for the car initial velocity $u = 0$ , acceleration $a = 5.0 {ms}^{- 2}$ time $t s$

For the truck Initial velocity $u = 22.0 m s^{- 1}$ for time $t = 10 s$ ,acceleration $a = - 1 {ms}^{- 2}$

The velocity gained by the car in $40 + 20 (t - 4)$ is given by the relation $v = u + at$ ,

Substituting the values, $v = 0 + 5 \times 4 = 20 {ms}^{- 1}$ ......................(1)

This is the constant velocity of the car after $4 s$ .

The distance covered by the car in $4 s e c$ is $= ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} 5 \times 4^{2} = 40 m$

As the total time of travel is $t$ , the car travelled for $4 s .$ with this speed. Therefore, the distance travelled by the car in $t \sec$ is given by

$s_{1} = 40 + 20 (t - 4) m$ .............................(2)

The truck is moving with the velocity of $22 {ms}^{_1}$ and slowed down with an acceleration of $- 1 {ms}^{- 2}$ ,

The distance travelled by the truck in first $10 \sec$ is $220 \times 10 = 220 m$ ...(3)

The distance travelled by the truck in next $(t - 10) \sec$ with a deceleration of $1 m s^{- 2}$ is given by $ut + \frac{1}{2} {at}^{2} = 22 (t - 4) - \frac{1}{2} 1 \times {(t - 10)}^{2}$

Therefore, total distance covered by the truck in $t$ sec. is given by

$220 + 22 (t - 10) - \frac{1}{2} (t - 10^{2})$ .................................(4)

When the car passes the truck, these distances should be equal i.e

$220 + 22 (t - 10) - \frac{1}{2} (t - 10^{2}) = 40 + 20 (t - 4)$

Or $220 + 22 t - 220 - \frac{1}{2} (t^{2} - 20 t + 100) = 20 t - 40$

$t^{2} + 24 t - 320 = 0$

On solving as quadratic expression , we get

, we get $t = 23.1 \sec$

The truck will cross the car after $23 \sec .$

Important Questions on Velocity and Acceleration

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