Question

A car accelerates from rest at a rate of 2.2 m/s^2 for 14.3s. The car then holds this speed for 16.0s, after which there is an acceleration of -3.0 m/s^2 until the car comes to rest.

a) Sketch a position vs. time graph for the car

b) Sketch a velocity vs, time graph for the car

c) What is the total distance traveled by the car?

Answer 1

Here is what I solved before, please modify the figures as per your question.. Ifthis helps then kindly rate 5-stars.

I'm going to split the problem into 3 parts.

acceleration,constant,deceleration

the final velocity of the acceleration part is the velocity of the constant part and the initial velocity of the deceleration part


We first find the distance the car traveled while accelerating

Vo=0 m/s (from rest)
a= 2 m/s^2
t= 10 s

Using the second kinematic equation
x=Vot +0.5at^2
= 0 +0.5*2*100
= 100 m

Now we find the final velocity

Using the first kinematic equation
V=Vo + at
=0+2*10
= 20 m/s

We find the distance the car has traveled while remaining at a constant speed
v = 20 m/s
t= 15 s
x= ?
x=vt
= 20*15 = 300 m

We find the distance the car has traveled while decelerating
Vo= 20 m/s
V= 0 m/s (the car stops)
a= -2.50 m/s^2
x= ?

Using the third kinematic equation
V^2 =Vo^2 +2ax
0=20^2 +2*-2.50*x
0= 400 + -5x
We move 400 to the left hand side of the equation
-400 = -5x
We divide both sides by -5
x= 80 m

We add the distances : 300 + 100 + 80 = 480 m

the car covered a total of 480 m