40.5k views
1 vote
A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average acceleration = 3.1 m/s2 for 7.9 s. What will be its velocity at that time.

User Plang
by
8.0k points

2 Answers

5 votes

Final answer:

The car's final velocity after accelerating from rest at 3.1 m/s^2 for 7.9 seconds will be 24.49 m/s.

Step-by-step explanation:

To calculate the velocity of a car that starts from rest and accelerates at a constant rate, you can use the formula:



v = u + at



where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.



In the case of the car starting from rest (u = 0.0 m/s) and accelerating at 3.1 m/s2 for 7.9 s, the formula simplifies to:



v = 0 + (3.1 m/s2)(7.9 s) = 24.49 m/s



So, the car's velocity after 7.9 seconds will be 24.49 m/s in the positive X-direction.

User Somesingsomsing
by
7.9k points
2 votes

Answer:

Final speed of the car, v = 24.49 m/s

Step-by-step explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration,
a=3.1\ m/s^2

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :


v=u+at


v=0+3.1* 7.9

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

User Loren Cahlander
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.