First slide

Rectilinear Motion

Question

In travelling a distance of 3 kilometre between points A and D, a car is driven at 100 kmh^-1 from A to B for t second and at 60 kmh^-1 from C to D for t second. If the brakes are applied for 4 second between B and C to give the car a uniform deceleration, the value of t is

Moderate

A

75.5 second
B

45.5 second
C

65.5 second
D

56.5 second

Solution

$\begin{array}{l} AB = ((100 \times \frac{5}{18}) t metre \\ CD = (60 \times \frac{5}{18}) t metre \\ For BC, v_{C} = v_{B} + at \\ \Rightarrow 60 \times \frac{5}{18} = 100 \times \frac{5}{18} + a (4) \\ \Rightarrow 4 a = - 40 \times \frac{5}{18} \\ \Rightarrow a = - \frac{50}{18} = - \frac{25}{9} {ms}^{- 2} \end{array}$
So the distance BC is
$\begin{array}{l} BC = 100 \times \frac{5}{18} \times 4 + \frac{1}{2} (- \frac{25}{9}) (4)^{2} \\ BC = \frac{1000}{9} - \frac{200}{9} = \frac{800}{9} m \\ However AB + BC + CD = 3000 m \\ \Rightarrow (160) (\frac{5}{18}) t + \frac{800}{9} = 3000 \end{array}$
$\begin{array}{l} \Rightarrow \frac{400}{9} t + \frac{800}{9} = 3000 \\ \Rightarrow 400 t + 800 = 27000 \\ \Rightarrow 400 t = 26200 \\ \Rightarrow t = 65 .5 s \end{array}$

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