On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Asked by Pragya Singh | 1 year ago | 63

The speed of car A = 36 km/h = 36 x (\( \dfrac{5}{8}\)) = 10 m/s

Speed of car B = 54 km/h = 54 x (\( \dfrac{5}{18}\)) = 15 m/s

Speed of car C = – 54 km/h = -54 x (\( \dfrac{5}{18}\))

= -15 m/s (negative sign shows B and C are in opposite direction)

Relative speed of A w.r.t C, V_{AC}= V_{A} – V_{B} = 10 – (-15) = 25 m/s

Relative speed of B w.r.t A, V_{BA} = V_{B} – V_{A} = 15 – 10 = 5 m/s

Distance between AB = Distance between AC = 1 km = 1000 m

Time taken by the car C to cover the distance AC, t = 1000/V_{AC} = \(
\dfrac{1000}{25}\) = 40 s

If a is the acceleration, then

s = ut + (\( \dfrac{1}{2}\)) at^{2}

1000 = (5 x 40) + (\( \dfrac{1}{2}\)) a (40) 2

a = \(
\dfrac{ (1000 – 200)}{800}\) = 1 m/s^{2}

Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s^{2}

The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

**(a)** \( x (t_2) = x (t_1) + v (t_1) (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a(t_2 – t_1)^2\)

**(b)** \( v(t_2) = v(t_1) + a(t_2 – t_1)\)

**(c)** Vaverage = \( \dfrac{ x(t_2) – x (t_1)}{(t_2 – t_1)}\)

**(d)** aaverage = \( \dfrac{v(t_2) – v (t_1) }{(t_2 – t_1)}\)

**(e)** \( x (t_2) = x (t_1) + v_{av} (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a_{av} (t_2 – t_1)^2\)

**(f)** \( x(t_2) – x (t_1)\) = Area under the v-t curve bounded by t- axis and the dotted lines.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

**(a)** t = 0 s to 10 s,

**(b)** t = 2 s to 6 s.

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of \( 15ms^{-1}\) and \(30ms^{-1}\)Verify that the graph shown in given figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = \( 10ms^{-2}\) Give the equations for the linear and curved parts of the plot.

On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h^{–1} (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^{–1}. For an observer on a stationary platform outside, what is the

**(a)** speed of the child running in the direction of motion of the belt ?.

**(b)** speed of the child running opposite to the direction of motion of the belt?

**(c)** time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^{-1}. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^{-1} and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?