Question

A car travels on the entrance ramp to a freeway as it accelerates from 30 mph to the freeway speed of 70 mph for 4.0 s. Suppose that the car moves with the constant acceleration.

a) Estimate the distance in feet that the car travels during this motion. (In feet)
b) During this motion what is the magnitude of the average acceleration of the car? (In ft/s^2)

Answer 1

Answer:

a) distance d = 293.36ft

b) acceleration a = 14.67ft/s^2

Explanation:

Acceleration is the change in velocity per unit time.

a = ∆v/t ....1

Given;

Initial velocity vi = 30mph × 5280ft/mile × 1/3600s/h

vi = 44ft/s

Final velocity vf = 70mph × 5280ft/mile × 1/3600s/h

vf = 102.67ft/s

time = 4.0s

From equation 1, acceleration is;

a = ∆v/t = (102.67-44)/4 = 14.67ft/s^2

Distance travelled can be given as;

d = ut + 0.5at^2 .....2

u = 44ft/s

t = 4

a = 14.67ft/s^2

Substituting into the equation 2

d = 44(4) + 0.5(14.67×4^2)

d = 293.36ft

Answer 2

Acceleration can be defined as the time rate of change in velocity. The acceleration of the given car is $\bold { 14.67\rm \ ft/s^2}$

Acceleration:

It can be defined by the time rate of change in velocity.

$a =\dfrac {v_f-v_i}t$

Where,

$v_i$- Initial velocity  = 30mph = 44ft/s

$v_f$ - Final velocity  = 70mph = 102.67 ft/s

$t$ - time = 4.0 s

Put the values in the first equation,

$a= \dfrac {102.67-44}{4 }\\\\a = 14.67\rm \ ft/s^2$

Therefore, the acceleration of the given car is $\bold { 14.67\rm \ ft/s^2}$.

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