Question

A body moving with constant acceleration covers between two points 60m apart in 6 sec. It's velocity as it passes the second point is 15m/s. A) What is its acceleration? B) What is its velocity at the first point?

Answer 1

The acceleration is `a in {0, 5/3}ms^-2` and the velocity `v_A in {15,5}ms^-1`

Explanation:

Let the `2` points be `A` and `B`

The distance `s=AB=60m`

The acceleration is `=ams^-2`

Let the velocities at `A` be `v_A` and at `B` be `v_B`

`v_A=?ms^-1`

`v_B=15ms^-1`

Therefore,

`a=(v_B-v_A)/(t)=(15-v_A)/(6)`

`6a=15-v_A`.......................`(1)`

and

`v_B^2=v_A^2+2as`

`15^2=v_A^2+2a*60`

`v_A^2+120a=225`..................`(2)`

Plugging in the value of `v_A` from equation `(1)` to equation `(2)`

`(15-6a)^2+120a=225`

`225-180a+36a^2+120a=225`

`36a^2-60a=0`

`3a^2-5a=0`

`a(3a-5)=0`

`a=0` or `a=5/3ms^-2`

So,

`v_A=15-6*5/3=15-10=5ms^-1`

or

`v_A=15-6*0=15ms^-1`

Answer 2

`A) 5/3`

`B) 5`

Explanation:

Part A requires the formula `vf^2 = vi^2 + 2ad`

You would have to manipulate the formula to get a:
`a = (vf^2 - vi^2)/ (2d)`

Taking vi from part B, you can fill in the equation.

`a = (15^2 - 5^2)/(2*60) = 5/3 m/s^2`

Part B requires the formula `d=((vf+vi)/2)t`

We know d = 60m, t = 6s, vf = 15, so we have to manipulate the formula to get vi:

`vi = (2d)/t - vf = (2*60)/6 - 15 = 5m/s`