Question

A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) find its original speed. (b) find its acceleration.

Answer 1

Speed equation: speed = total distance/total time

speed = 40.0m / 8.50s

speed = 4.71 m/s

acceleration equation: acceleration = change in velocity/time

4.71 - 2.80 = 1.91 (change in velocity)

acceleration = 1.91/8.50

acceleration = .22 m/s^2

a) 4.71 m/s
b) .22 m/s^2

Answer 2

Answer:

For a: The original speed of the truck is 4.76 m/s

For b: The acceleration of the truck is $-0.23m/s^2$

Explanation:

• For a:

Speed is defined as the rate at which an object moves with respect to time.

To calculate the time taken for the given speed, we use the equation:

$s=\frac{d}{t}$

where,

s = speed of the truck

d = distance traveled = 40.0 m

t = time taken by truck = 8.50 s

Putting values in above equation, we get:

$s=\frac{40.0m}{8.50s}=4.76m/s$

Hence, the original speed of the truck is 4.76 m/s

• For 2:

Acceleration is defined as the rate of change of velocity with respect to time.

Mathematically,

$a=\frac{v-u}{t}$

where,

v = final velocity  of the truck = 2.80 m/s

u = initial velocity  of the truck = 4.76 m/s

t = time taken  = 8.50 s

Putting values in above equation, we get:

$a=\frac{2.80-4.76}{8.50}=-0.23m/s^2$

Negative sign represents slowing down or deceleration.

Hence, the acceleration of the truck is $-0.23m/s^2$