Question

Two cars are travelling towards one another on the two sides of a long straight road. Each car stops when its speed is

0 m s^{- 1}

. The time, in seconds, after the first car starts to move is

t

. The acceleration,

a_{1} m s^{- 2}

, of the first car is given by

a_{1} = 5 - 2 t

. The maximum velocity that the first car achieves is

26.01 m s^{- 1}

.
The velocity,

v_{2} m s^{- 1}

, of the second car is given by

v_{2} = t^{2} - 16

.
How far does the second car travel?

Accepted Answer

Given: Two cars are travelling towards one another on the two sides of a long straight road. Each car stops when its speed is $0 m s^{- 1}$ . The time, in seconds, after the first car starts to move is $t$ . The acceleration, $a_{1} m s^{- 2}$ , of the first car is given by $a_{1} = 5 - 2 t$ . The maximum velocity that the first car achieves is $26.01 m s^{- 1}$
The velocity of the second car is given by:

$v_{2} = t^{2} - 16$

To find the time when the first car stops moving is given by:

We find out the expression of the velocity, using,

$a = \frac{d v}{d t} \Rightarrow d v = a d t \Rightarrow \int_{u}^{v} d v = \int_{0}^{v} a d t \Rightarrow v - u = \int_{0}^{t} (5 - 2 t) d t \Rightarrow v = - (t^{2}) + 5 (t) + u . . . . . . . . . . . . . . . . . . . . . . (1)$

For velocity to be maximum, we have,

$\frac{d v}{d t} = 0 \Rightarrow a_{1} = 0 \Rightarrow 5 - 2 t = 0 \Rightarrow t = 2.5 s$

Putting the values in the velocity expression of equation $(1)$ , we get,

$v = - (t^{2}) + 5 (t) + u \Rightarrow 26.01 = - {(2.5)}^{2} + 5 (2.5) + u \Rightarrow u = 19.76 {ms}^{- 1}$

We find out the expression of the velocity obtained previously is:

$v = - (t^{2}) + 5 (t) + 19.76 . . . . . . . . . . . . . . . . . . . . . . (1)$

For velocity to be zero, we have,

$0 = - (t^{2}) + 5 (t) + 19.76 \Rightarrow (t^{2}) - 5 (t) - 19.76 = 0$

Solving the quadratic equation, we get,

$t = \frac{5 \pm \sqrt{5^{2} - 4 (1) (19.76)}}{2} = 7.53 s$

The time taken by the first car to stop $= 7.52 s$

The displacement of the second car during the time interval is given by:

$\int_{0}^{s} d x = \int_{0}^{7.52} (t^{2} - 16) d t \Rightarrow \int_{0}^{s} d x = \frac{1}{3} \int_{0}^{7.52} (t^{2}) d t - \int_{0}^{7.52} 16 d t \Rightarrow s - 0 = 0.33 (7 . 52^{3} - 0) - 16 (7.52 - 0) \Rightarrow s = 20 m$

Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 8: EXERCISE 6D

Jan Dangerfield Mathematics Solutions for Exercise - Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 8: EXERCISE 6D

Questions from Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 8: EXERCISE 6D with Hints & Solutions

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