Question

A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the traveled distance of the car compared to the truck?
A. Half as much
B. The same
C. Twice as much
D. Four times as much
E. One quarter as much

Answer 1

The traveled distance of the car is four times as much compared to the truck

Further explanation

Acceleration is rate of change of velocity.

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

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

Initial velocity of Car = Initial velocity of Truck = u = 0 m/s

Acceleration of Car = Acceleration of Truck = a

Time Taken of Truck = t

Time Taken of Car = 2t

Unknown:

The Travel Distance of the Car : The Travel Distance of the Truck = ?

Solution:

We will compare the travel distance of the car and truck

$d_{car} : d_{truck} = (ut_{car} + \frac{1}{2}at_{car}^2) : (ut_{truck} + \frac{1}{2}at_{truck}^2)$

$d_{car} : d_{truck} = (0(t_{car}) + \frac{1}{2}at_{car}^2) : (0(t_{truck}) + \frac{1}{2}at_{truck}^2)$

$d_{car} : d_{truck} = \frac{1}{2}at_{car}^2 : \frac{1}{2}at_{truck}^2$

$d_{car} : d_{truck} = \frac{1}{2}at_{car}^2 : \frac{1}{2}at_{truck}^2$

$d_{car} : d_{truck} = t_{car}^2 : t_{truck}^2$

$d_{car} : d_{truck} = (2t)^2 : (t)^2$

$d_{car} : d_{truck} = 4t^2 : t^2$

$d_{car} : d_{truck} = 4 : 1$

$d_{car} = 4 d_{truck}$

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

Answer 2

The traveled distance of the car compared to the truck will be four times as much.

Let,

The initial velocity at rest will be:

• v m/s (0 m\s)

Acceleration,

• a m/s

Time of acceleration of truck,

• t sec

Time of acceleration of car,

• 2t sec

The equation of the motion,

→ $S = vt +\frac{a}{t} at^2$

By putting the values, we get

→ $S = \frac{1}{2} at^2$

hence,

The distance travelled by car will be:

→ $S' = v\times 2t+\frac{1}{2}\times a\times (2t)^2$

→     $= 2vt+\frac{1}{2}\times a\times 4t^2$

→     $= 0 +4(\frac{1}{2}at^2 )$

→     $= 4(\frac{1}{2} at^2)$

→     $=4$

Thus option D is the correct alternative.

Learn more:

https://brainly.com/question/18650822