Question

In a test run, a certain car accelerates uniformly from zero to 24.0 m/s in 2.95 s. (a) What is the magnitude of the car’s acceleration? (b) How long does it take the car to change its speed from 10.0 m/s to 20.0 m/s? (c) Will doubling the time always double the change in speed? Why?

Answer 1

(a) 8.14 m/s^2

Step-by-step explanation:

Acceleration of the car is given by:

`a=(v-u)/(t)`

where

u = 0 is the initial velocity

v = 24.0 m/s is the final velocity

t = 2.95 s is the time taken

Substituting numbers into the equation, we find

`a=(24.0 m/s-0)/(2.95 s)=8.14 m/s^2`

(b) 1.23 s

Step-by-step explanation:

We can use the same formula for the acceleration:

`a=(v-u)/(t)`

where this time, we have

u = 10.0 m/s

v = 20.0 m/s

a = 8.14 m/s^2

Substituting numbers into the equation, we can find the time taken, t:

`t=(v-u)/(a)=(20.0 m/s-10.0 m/s)/(8.14 m/s^2)=1.23 s`

c) If the acceleration is constant, yes

Step-by-step explanation:

Let's take again the equation for the acceleration:

`a=(v-u)/(t)`

We can rewrite it as:

`(v-u) = at`

where the term (v-u) represents the change in velocity. From the formula we see that, if a (acceleration) is constant, then (v-u) is directly proportional to the time t: therefore, if t doubles, the change in velocity (v-u) doubles as well.