Question

A car travels on a straight, level road.(a) Starting from rest, the car is going 36 ft/s (25 mi/h) at the end of 5.5 s. What is the car's average acceleration in ft/s2?magnitude______ ft/s2What direction? in the direction of motion or opposite to the direction of motion? (b) In 4.5 more seconds, the car is going 72 ft/s (49 mi/h). What is the car's average acceleration for this time period?magnitude_______ft/s2What direction? in the direction of motion or opposite to the direction of motion? (c) The car then slows to 54 ft/s (37 mi/h) in 3.5 s. What is the average acceleration for this time period?magnitude_______ft/s2What direction? in the direction of motion or opposite to the direction of motion? (d) What is the overall average acceleration for the total time?magnitude________ft/s2What direction? in the direction of motion or opposite to the direction of motion?

Answer 1

a) Acceleration, a = change in velocity/time = (v - u)/t

where

u is the initial velocity

v is the final velocity

t = time

From the information given,

u = 0 because it started from rest

v = 36

t = 5.5

Thus,

a = (36 - 0)/5.5 = 36/5.5 = 6.54 ft/s^2

The car's average acceleration = 6.54 ft/s^2

The direction is in the direction of motion

b) From the information given,

t = 4.5

u = 36

v = 72

a = (72 - 36)/4.5 = 8 ft/s^2

The car's average acceleration for this time period is 8ft/s^2

The direction is in the direction of motion

c) From the information given,

u = 72

v = 54

t = 3.5

a = (54 - 72)/3.5 = - 5.14 ft/s^2

The car's average acceleration for this time period is - 5.14ft/s^2

The direction is in the opposite direction of motion

d) Average acceleration = total acceleleration/3 = (6.54 + 8 - 5.14)/3

Average acceleration = 3.13 ft/s^2

The direction is in the direction of motion