A truck covers 80 m in 17 s while smoothly slowing down to a final speed

[r-soy]

A truck covers 80 m in 17 s while smoothly slowing down to a final speed of 5.6 m/s
a ) find its original speed
b) Final its acceleration

I want check my answer :
a )

Dx = 80 m
t = 17 s
v = 5.6 m/s
v0= ??

by appling the rule

Dx = 1/2(v+v0)t
80 = 1/2(5.6 + v0) 17
v0 = .5 X 80 - 5.6 /17
=2.023 m/s

-------------

b)

v = v0 + at
5.6 = 2.023 + 17a
a = 5.6 -2.023/17 = 0.214

plese help me and wha thw unit of a will be ?

 

[sjb-2812]

Something seems wrong, as slowing down implies, well, that the initial speed was faster than the final. But 2.0 m/s is slower than 5.6 m/s.

If the speed is in m/s, and time in s what is the unit of acceleration? What is acceleration?

 

[r-soy]

how ??

 

[r-soy]

plese I want your help

 

[r-soy]

now I try to sovle this queation by another way anf please help me If it correct or not



Dx = 1/2(v + v0)t

80=1/2(5.6+v0) 17

80 = 1/2(8.6 + 17v0)

80 = 4.25 + 17v0

v0 = 75.75/17 =4.45



plese check my answer .

 

[Mark44]

now I try to sovle this queation by another way anf please help me If it correct or not



Dx = 1/2(v + v0)t

80=1/2(5.6+v0) 17

80 = 1/2(8.6 + 17v0)

Where did 8.6 come from? Also, you multiplied v0 by 17, but you didn't also multiply 5.6.

80 = 4.25 + 17v0

v0 = 75.75/17 =4.45



plese check my answer .

Please check that you have written the problem correctly. As given in the first post, it's not possible for the truck to slow down to 5.6 m/sec from a higher speed, and cover 80 m.

Think about it this way: Suppose the truck was moving at a constant speed of 5.6 m/sec. During the 17 seconds, the truck would have traveled 5.6 m/sec * 17 sec = 95.2 m.

On the other hand, if the truck started at a higher speed and slowed to 5.6 m/sec, it would have covered more than 95.2 m in the 17 seconds.

 

[Mark44]

r-soy,
Are you going to follow up on this problem?