A bus starts to move from rest. If it is accelerated by 0.8m/s2, calculate the velocity and distance traveled after 8 s.
1 answer:
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Answer:
(a)Therefore the highest altitude attained by the object is =576 ft .
(b)Therefore the object takes 6 sec to fall to the ground.
Explanation:
Initial velocity: Initial velocity is a velocity from which an object starts to move.
u is usually used for notation of initial notation.
Final velocity: Final velocity is a velocity of an object after certain second from starting.
The final velocity is denoted by v.
Acceleration: The difference of final velocity and initial velocity per unit time
The S.I unit of acceleration is m/s².
(a)
Given that u= 128 ft\sec and g = 32 ft/sec².
At highest point the velocity of the object is 0 i.e v=0
Since the displacement is opposite to the gravity.
Therefore acceleration( a)= -g = -32 ft/sec².
To find the time this to happen we use the following formula
Here v=0
⇒0=128+(-32) t
⇒32t=128
⇒t = 4 sec
To determine the height we use the following formula
⇒s= 256 ft
Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .
(b)
At the highest point the velocity of the object is 0.
so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft
To determine the time to fall we use the following formula
⇒t=6 sec
Therefore the object takes 6 sec to fall to the ground.