A motorcyclist's velocity is initially 20m/s. If it maintains a constant acceleration of 4m/s/s, what distance will it cover in 15s?
2006-11-01 09:37:51 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
d = ut +1/2at^
where
d = distance travelled
u = initial velocity
a = acceleration
t = time
d = (20 x 15) + [(4 x 15^)/2]
d = 300 + [900/2]
d = 300 + 450
d = 750m
2006-11-01 23:56:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Final Velocity = 20 + 4 * 15 = 80 m/s
mean Velocity = (20 + 80)/2 = 50 m/s
distance = 50 * 15 = 750 m
2006-11-01 10:56:54 · answer #2 · answered by Ormoz 3 · 0⤊ 0⤋
s = ut + 1/2ft^2
where:
s = distance
u = initial velocity
t = time
f = acceleration
so:
s = 20 * 15 + ((4 * 15^2) / 2)
s = 750m
2006-11-01 09:54:22 · answer #3 · answered by Sam 3 · 0⤊ 0⤋
using the formula
s=ut + 1/2at^2 where s= distance, u=initial vel.,t=time, a=acceleration
s=20x15 + 1/2x4x15^2
s=300+450 =750m
2006-11-01 09:55:16 · answer #4 · answered by obiora c 2 · 0⤊ 0⤋
For constant acceleration (a):
d=(init v)( t) + 0.5at^2 = 20x15 + 0.5x4x225 = 750 m
If this guy was being chased by cops, he'd be gone in a blink of an eye !!!!!!!!!!!!!!!!!!!!
2006-11-01 09:39:57 · answer #5 · answered by Mech_Eng 3 · 0⤊ 0⤋
I don't know the distance it will cover,but if it's my bike it'll all be done on the back wheel.
2006-11-01 09:41:09 · answer #6 · answered by jixer 3 · 0⤊ 0⤋