Question

A particle accelerates from rest with an Acceleration of 10m/s^2 and then Decelerates at a rate of 15m/s^2. If the particle was in motion for a total time of 10s then find the maximum velocity of the particle.
(A) 50 m/s
(B) 60m/s
(C) 20 m/s
(D) 25m/s​

Answer 1

ANSWER :

✴ Given

→ A particle accelerates from rest with an acceleration of 10m/s$^2$ and then decelerates at a rate of 15m/s$^2$

→ Time interval = 10s

✴ To Find :

→ Maximum velocity of the particle.

✴ Short method :

→ A body accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is given by

• Vmax = αβt/α+β

✴ Calculation :

▪ α = 10m/s$^2$

▪ β = 15m/s$^2$

▪ t = 10s

⏭ Vmax = 10×15×10/10+15

⏭ Vmax = 1500/25

⏭ Vmax = 60m/s

▶ Maximum speed of the particle = 60mps (Option : B)

Answer 2

Answer:

A particle accelerates from rest with an acceleration of 10m/s and then decelerates at a rate of 15m/s

Time interval = 10s

To Find :

Maximum velocity of the particle.

A body accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is given by

Vmax = αβt/α+β

Calculation :

α = 10m/s

β = 15m/s

t = 10s

Vmax = 10×15×10/10+15

Vmax = 1500/25

Vmax = 60m/s

Maximum speed of the particle = 60mps (Option : B)