Advertisements
Advertisements
प्रश्न
A cyclist driving at 36 kmh−1 stops his motion in 2 s, by the application of brakes. Calculate
- retardation
- distance covered during the application of brakes.
Advertisements
उत्तर
Initial velocity = u = 36 kmh−1
= u = `36xx5/18` ms−1 = 10 ms−1
Final velocity = v = 0
Time = t = 2s
(i) Retardation = a = ?
v = u + at
0 = 10 + a (2)
2a = −10
a = `(-10)/2` = −5 ms−2
(ii) Distance covered S =?
v2 − u2 = 2aS
(0)2 − (10)2 = 2(−5) S
−10S = −100
S = `100/10`
S = 10 m
APPEARS IN
संबंधित प्रश्न
State the type of motion represented by the following sketches in Figures.
Give an example of each type of motion.
A car travels with uniform velocity of 25 m s-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find:
- The distance which the car travels before the brakes are applied,
- Retardation and
- The distance travelled by car after applying the brakes.
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/s. It took 2 seconds for it to go from B to C. What is the acceleration of the ball as it goes from B to C?
When is the acceleration due to gravity negative?
From the diagram given below, calculate deceleration.
A body falls towards the earth. Does it have positive or negative acceleration?
The initial velocity of a car is 10 ms-1. It moves with an acceleration of 2 ms-2. What will be its speed after 10 seconds?
State if the following situation is possible:
A body moving with a constant velocity in an accelerated motion.
If the velocity of a body does not change, then its acceleration is ______.
The acceleration of the body that moves with a uniform velocity will be ______.
