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3 years ago

A truck accelerates to a velocity of 38 m/s over 755 m of road

Physics

1 answer:

den301095 [7]3 years ago

6 0

1) The acceleration is $0.96 m/s^2$

2) The time taken is 39.6 s

Explanation:

Since the motion of the truck is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the truck in this problem,

u = 0 (it starts from rest)

v = 38 m/s

s = 755 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{38^2-0}{2(755)}=0.96 m/s^2$

For this part we can use the following suvat equation

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken for the velocity to change from u to v

In this problem,

u = 0

v = 38 m/s

$a=0.96 m/s^2$

Solving for t,

$t=\frac{v-u}{a}=\frac{38-0}{0.96}=39.6 s$

Learn more about accelerated motion:

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