Home
Class 12
PHYSICS
A particle starts from rest, accelerates...

A particle starts from rest, accelerates at `2(m)/(s^2)` for 10 s and then goes for constant speed for 30 s and then decelerates at `4(m)/(s^2)`. Till it stops. What is the distance travelled by it?

A

750 m

B

800 m

C

700 m

D

850 m

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will break down the motion of the particle into three distinct parts: acceleration, constant speed, and deceleration. ### Step 1: Calculate the distance during acceleration The particle starts from rest and accelerates at \(2 \, \text{m/s}^2\) for \(10 \, \text{s}\). 1. **Initial velocity (u)**: \(0 \, \text{m/s}\) (starts from rest) 2. **Acceleration (a)**: \(2 \, \text{m/s}^2\) 3. **Time (t)**: \(10 \, \text{s}\) Using the formula for final velocity: \[ v = u + at \] Substituting the values: \[ v = 0 + (2 \, \text{m/s}^2)(10 \, \text{s}) = 20 \, \text{m/s} \] Now, calculate the distance traveled during this acceleration using: \[ s = ut + \frac{1}{2} a t^2 \] Substituting the values: \[ s = 0 \cdot 10 + \frac{1}{2} \cdot 2 \cdot (10)^2 = \frac{1}{2} \cdot 2 \cdot 100 = 100 \, \text{m} \] **Distance during acceleration**: \(100 \, \text{m}\) ### Step 2: Calculate the distance during constant speed The particle moves at a constant speed of \(20 \, \text{m/s}\) for \(30 \, \text{s}\). Using the formula for distance: \[ s = vt \] Substituting the values: \[ s = 20 \, \text{m/s} \cdot 30 \, \text{s} = 600 \, \text{m} \] **Distance during constant speed**: \(600 \, \text{m}\) ### Step 3: Calculate the distance during deceleration The particle decelerates at \(4 \, \text{m/s}^2\) until it stops. 1. **Initial velocity (u)**: \(20 \, \text{m/s}\) (the speed at which it was traveling before deceleration) 2. **Final velocity (v)**: \(0 \, \text{m/s}\) (it stops) 3. **Deceleration (a)**: \(-4 \, \text{m/s}^2\) (negative because it's deceleration) Using the formula: \[ v^2 = u^2 + 2as \] Rearranging for distance \(s\): \[ s = \frac{v^2 - u^2}{2a} \] Substituting the values: \[ s = \frac{0 - (20)^2}{2 \cdot (-4)} = \frac{-400}{-8} = 50 \, \text{m} \] **Distance during deceleration**: \(50 \, \text{m}\) ### Step 4: Calculate the total distance traveled Now, we sum up all the distances from each part of the motion: \[ \text{Total distance} = \text{Distance during acceleration} + \text{Distance during constant speed} + \text{Distance during deceleration} \] \[ \text{Total distance} = 100 \, \text{m} + 600 \, \text{m} + 50 \, \text{m} = 750 \, \text{m} \] ### Final Answer The total distance traveled by the particle is \(750 \, \text{m}\). ---
To solve the problem step by step, we will break down the motion of the particle into three distinct parts: acceleration, constant speed, and deceleration. ### Step 1: Calculate the distance during acceleration The particle starts from rest and accelerates at \(2 \, \text{m/s}^2\) for \(10 \, \text{s}\). 1. **Initial velocity (u)**: \(0 \, \text{m/s}\) (starts from rest) 2. **Acceleration (a)**: \(2 \, \text{m/s}^2\) 3. **Time (t)**: \(10 \, \text{s}\) ...
Promotional Banner

Similar Questions

Explore conceptually related problems

A car, starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f/ 2 to come to rest. If the total distance travelled is 15 s, then

A car, starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f/ 2 to come to rest. If the total distance travelled is 15 s, then

A car , starting from rest, accelerates at the rate f through a distance S then continues at constant speed for time t and then decelerates at the rate (f)/(2) to come to rest . If the total distance traversed is 15 S , then

A bus starting from rest, accelerates at the rate of f through a distance S, then continues at constant speed for time t and declerates at the rate f/2 to come to rest. If the total distance travelled is 15 S then S =

A particle starts from rest and accelerates constantly with 2m//s^(2) for 10 s and then retards uniformly with same rate till it comes to rest. Draw the x-t, v-t and a-t graphs.

A car, starting from rest, accelerates at the rate (f) through a distance (S), then continues at constant speed for some time (t) and then decelerates at the rate f//2 to come to rest. If the total distance is 5 S , then prove that S=1/2/ ft^(2) .

A body starts from rest and accelerates with 4 "m/s"^2 for 5 seconds. Find the distance travelled with 5 seconds.

A particle starts from rest and has an acceleration of 2 ms^(–2) for 10 s. After that, it travels for 30 s with constant speed and then undergoes a retardation of 4 ms^(–2) and comes to rest. Calculate the total distance covered by the particle.

A particles starts from rest and has an acceleration of 2m//s^(2) for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of 4m//s^(2) and comes back to rest. The total distance covered by the particle is

A train starts from rest and accelerates uniformly at a rate of 2 m s^(-2) for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find : (i) the maximum velocity reached, (ii) the retardation in the last 50 S, (iii) the total distance travelled, and (iv) the average velocity of the train.