Physics

1 -coming to rest after another 20.0 seconds. Calculate 

a) the acceleration on the car during ea. A car starts from rest and accelerates uniformly for 10.0s, reaching a speed of 20.0m/s. It then continues traveling at that speed for 30.0 additional seconds before decelerating uniformly and ch of the three intervals 

a1=Vf-Vi/t=20.0m-0/s/10.0m/s=2.00m/s^2

Because the car continues at the same speed Vf = Vi at t=30.0s

a2=Vf-Vi/t=20.m/s-20.0m/s/30.0s=0m/s^2

a3=Vf-Vi/t= 0-20.0m/s/20.0s=-1m/s^2

b) And the distance traveled during each time

d1=Vi1*t1+ 0.5 *at1^2=0+ 0.5*2.00m/s^2* (10.0s) ^2=100.m

d2=Vi2*t2 +0.5*at2^2=20.0m/s*30.0s + 0.5*0m/s^2*30.0s=600.m

d3=Vi3*t3 +0.5at3^2=20.0m/s*20.0s +0.5*(-1m/s^2)*(20.0s) ^2=200.m

2. A ball is dropped from the roof of a 98.0m building (in a vacuum). Calculate 

a) how long it takes to reach the ground

Calculation of time using formulas: d=Vit+0.5gt^2

-98= 0 + 0.5*(-9.8m/s^2)*t^2 => t=4.47s

b) And what its velocity is just before it hits the ground.

Vf^2=Vi^2 +2g*d = 0+2(-9.8m/s^2)* -98.0m => Vf=43.8m/s @ -90.0⁰

3. A ball is thrown straight up from the edge of the same building with an initial velocity of 20.0m/s. Calculate 

a) how high it will rise 

b) how long it will take to reach that peak height 

c) how long it will take to return to the same height as the edge of the roof 

d) its velocity at that point 

e) how long it will take to hit the ground 

f) and its velocity just before hitting the ground

Submit as Assign2B.