A car travels a distance of d=25.5m in the positive x-direction in a time of t1=

Question

A car travels a distance of d=25.5m in the positive x-direction in a time of t1=24.5s. The car immediately brakes and comes to a rest in t2=7s

a) What was the car's average velocity in the horizontal direction, in meters per second, during t1?

b) What was the acceleration, in meters per second squared, during time interval t1, assuming the car started from rest and moved with a constant acceleration.

c) What was the car's instantaneous velocity in the horizontal direction, in meters per second, when it began braking?

d) Using the result from part c), what was the car's horizontal component of acceleration, in meters per second squared, during the braking period?

Homework 1 Begin Date: 1/15/2018 12:00:00 AM -- Due Date: 1/29/2018 11:59:00 PM End Date: 1/29/2018 11:59:00 PM (6%) Problem 15: A car travels a distance d = 25.5 m in the positive x-direction in a time of t1 = 24.5 s. The car immediately brakes and comes to rest in t2 = 7 s Randomized Variables d = 25.5 m t1 24.5 s ©theexpertta.com 25% Part (a) What was the car's average velocity in the horizontal direction, in meters per second, during t? × 25% Part (b) what was the acceleration, in meters per second squared, during time interval tl , assuming the car started from rest and moved with a constant acceleration? Grade Summary Deductions Potential a 0.042441 0% 100% Submissions Attempts remaining: 6 (0% per attempt) detailed view tan() ( 789 HOME cosO cotan0 asin) acos0 atan)acotan() sinh0 0% 0 coshO tanhO cotanh0 Degrees O Radians END ) BACKSPACE E CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 1 Feedback: 2% deduction per feedback 25% Part (c) What was the car's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 25% Part (d) Using the result from part (c), what was the car's horizontal component of acceleration, in meters per second squared, during the braking period?

Explanation / Answer

ANs:- given data d = 25.5m

t1 = 24.5s

t2= 7s

Vave = total displacement /total time

                = -25.5/7-24.5

                = 1.46m/s

B ]vf = vi +at

7.29 = 0 + a * 24.5

a=0.297m/s^2

c] use kinematical equation

s = ½ (vi + vf )t

25.5 = ½ (Vi + 0 )7

25.5 = 3.5Vi

Vi = 7.29m/s

D]

Vf = Vi +at

0 = 7.29 +a *7

a= -1.04m/s^2

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