Name: Relative Velocity in One Dimensional Motion
Uploaded: 2019-07-19
Duration: 8 min 17 s
Description: Rest and motion always go together. E.g., the Earth is at rest for us, but for a person in space, it’s revolving around the sun. We will explore relative mot...

Question

A motor car moving at a speed of

72 km / h

cannot come to stop in less than

3.0 s

while for a truck this time interval is

5.0 s

. On a highway, the car is behind the truck both moving

72 km / h

. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck? Human response time is

0.5 s

.

Accepted Answer

For truck initial velocity: $u = 72 \times \frac{5}{18} m / s = 20 m / s$

The final velocity of truck will be $v = 0$ and the time taken by truck to stop is $0.5 s$ .

By using equation of motion: $v = u + a t$

$\Rightarrow 0 = 20 m s^{- 1} + a \times (5 s)$

$a_{t} = \frac{- 20 m s^{- 1}}{5 s} = - 4 m s^{- 2}$

For car the initial velocity is $u = 72 km h^{- 1} = 20 m s^{- 1}$ and to get to rest it takes time $t = 3 s$ .

By using equation of motion: $v = u + a t$

$\Rightarrow 0 = 20 m s^{- 1} + a_{c} (3 s) \Rightarrow a_{c} = - \frac{20}{3} m s^{- 2}$

Let car is at distance $x$ from the truck and $t$ be the time to cover this distance. Since, the time of response of human eye is $0.5 s$ .
Then the real time to see and react will be $(t - 0.5 s)$ .

So, velocity of car after $t$ will be $v_{c} = u + a_{c} t$

$v_{c} = 20 m s^{- 1} + (- \frac{20}{3} m s^{- 2}) (t - 0.5 s)$

Velocity of truck after time $t$ is $v_{t} = u + a t$

$v_{t} = (20 m s^{- 1}) + (- 4 m s^{- 2}) t$

Since, before getting collide they will stop i.e. $v_{t} = v_{c}$

$20 m s^{- 1} - (\frac{20}{3} m s^{- 2}) (t - 0.5 s) = 20 m s^{- 1} - (4 m s^{- 2}) t$

$\Rightarrow t = (\frac{1}{4 m s^{- 2}}) \times (\frac{20}{3} m s^{- 2}) \times (t - 0.5 s)$

$\Rightarrow t = \frac{5}{3} \times (t - 0.5 s) \Rightarrow 2 t = 2.5 s \Rightarrow t = 1.25 s$

So, the distance travelled by truck in this time is $s_{t} = u t + \frac{1}{2} a_{t} t^{2}$

$\Rightarrow s_{t} = (20 m s^{- 1}) (1.25 s) + \frac{1}{2} (- 4 m s^{- 2}) {(1.25 s)}^{2}$

$\Rightarrow s_{t} = 25 m - 3.125 m = 21.875 m$

So, the distance travelled by car in this time is $s_{c} = u (0.5 s) + u (t - 0.5 s) + \frac{1}{2} a_{c} {(t - 0.5 s)}^{2}$

$\Rightarrow s_{c} = (20 m s^{- 1} \times 0.5 s) + (20 m s^{- 1}) (1.25 s - 0.5 s) + \frac{1}{2} (- \frac{20}{3} m s^{- 2}) {(1.25 s - 0.5 s)}^{2}$

$\Rightarrow s_{c} = 10 m + 15 m - 1.875 m = 23.125 m$

So, the distance between the car and truck is $23.125 m - 21.875 m = 1.25 m$

Hence, the car should maintain at least $1.25 m$ of distance from the truck to avoid collision.

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