Acceleration Question

[jtagtp]

An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?


I know how to find the Vel. from this (64m/4s) but i dont understand how im able to find the acceleration?

The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

J

[arildno]

1.Your expression for the (instantaneous) velocity is totally wrong; you're mixing up with the average velocity.

2. Ask yourself:
How is "distance travelled in time t" related to constant acceleration?

3. Welcome to PF!

[97gtpacecar]

Man im so lost. Ok i relized i used the average velocity, but what i need to do is use the instantaneous velocity? I never did understand inst. vel?

Thanks for welcoming me, i loved physics in high school but so far this college physics is :confused: :mad: :cry:

[arildno]

OK, let's take it slow&easy:
What formulas do you know which is about constant acceleration?

[97gtpacecar]

constant acceleration =(v(t2)-v(t1))/(t2-t1) = chance in v over the change in time.

[HallsofIvy]

Do you know d= (1/2)a t2 for constant acceleration starting from rest?

[97gtpacecar]

So i guess i can try and use the formula and switch it around. a=2d/(t^2)

[97gtpacecar]

ummm but the problems is that this in not from a dead stop.

[needhelpperson]

An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?


I know how to find the Vel. from this (64m/4s) but i dont understand how im able to find the acceleration?

The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

J

Either the answer in the book is wrong, or the question is flawed.

we know that vi = 0

so vf/2 = 64/8 = 16m/s

a = 16m/s/8s = 2m/s^2. If the acceleration is constant then it should be 2m/s^2 the whole time. However, i don't know where the 1.6m/s^2 came from, unless it didn't start from rest....

[97gtpacecar]

Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t

[needhelpperson]

Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t


read it once more....

An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line....

and besides if it's constant acceleration, it should be the same in the second section.