Two cars P and Q are travelling on a straight path and are $60$ m apart as shown in the figure; Car P is moving with a constant velocity of $36$ kmph, while car Q is moving at a constant velocity of $18$ kmph. At this instant, the driver in car P applies the brake and collision occurs with car Q after $30$ seconds. Assuming uniform deceleration due to braking, which one of the following is the CORRECT velocity (in m/s) of the car P just before the collision?
To solve this problem, we need to determine the velocity of car P just before the collision. Let's break down the steps:
- Convert the velocities from km/hr to m/s.
- Velocity of car P: \(36 \, \text{kmph} = \frac{36 \times 1000}{3600} = 10 \, \text{m/s}\)
- Velocity of car Q: \(18 \, \text{kmph} = \frac{18 \times 1000}{3600} = 5 \, \text{m/s}\)
- Calculate the relative velocity of car P with respect to car Q.
- Relative velocity, \(v_{\text{rel}} = v_P - v_Q = 10 - 5 = 5 \, \text{m/s}\)
- Determine the relative distance to be closed for a collision to occur.
- Relative distance = 60 m
- Use the formula for relative motion under uniform acceleration:
- The time taken for collision, \(t = 30 \, \text{s}\).
- Apply the equation of motion for car P:
- The equation \(v = u + at\) can be rearranged to find final velocity \(v\) just before collision.
- Let \(a\) be the deceleration. From \(s = ut + \frac{1}{2}at^2\), where \(s = 60 \, \text{m}\) and \(u = \text{initial relative velocity} = 5 \, \text{m/s}\),
- \(60 = 5 \times 30 + \frac{1}{2} \times a \times (30)^2 \Rightarrow 60 = 150 + 450a \Rightarrow a = -0.2 \, \text{m/s}^2\)
- Using \(v = u + at\) for car P:
- \(v = 10 + (-0.2) \times 30 = 4 \, \text{m/s}\)
Thus, the velocity of car P just before the collision is 4 m/s.
Important Questions from Speed Distance and Time
- Two cars, P and Q, start from a point X in India at 10 AM. Car P travels North with a speed of 25 km/h and car Q travels East with a speed of 30 km/h. Car P travels continuously but car Q stops for some time after travelling for one hour. If both the cars are at the same distance from X at 11:30 AM, for how long (in minutes) did car Q stop?
In a 500 m race, P and Q have speeds in the ratio of 3: 4. Q starts the race when P has already covered 140 m.
What is the distance between P and Q (in m) when P wins the race?A vehicle is moving at a speed of 12 m/s on a level road. It applies emergency brakes and starts to skid without rolling in a straight path. The deceleration of the vehicle is constant after braking and it comes to rest at a distance of 15 m. Assuming, $g = 10$ m/s$^2$, the coefficient of kinetic friction between the tyres and road is _________ [round off to 2 decimal places]
- The distance between Delhi and Agra is 233 km. A car P started travelling from Delhi to Agra and another car Q started from Agra to Delhi along the same road 1 hour after the car P started. The two cars crossed each other 75 minutes after the car Q started. Both cars were travelling at constant speed. The speed of car P was 10 km/hr more than the speed of car Q. How many kilometers the car Q had travelled when the cars crossed each other?
Two trains started at 7AM from the same point. The first train travelled north at a speed of 80km/h and the second train travelled south at a speed of 100 km/h. The time at which they were 540 km apart is _______________ AM.
