65 Views

a cyclist riding with a speed of 27kmph.As he approaches a circular turn on the road of radius 80m, he applies breaks and reduces his speed at the constant rate of 0.50m/s every second. the net acceleration of cyclist on the circular turn is


narendragunapati 28th Sep, 2020
Answer (1)
ARYAN SAGAR 28th Sep, 2020
Dear student,

Due to Braking,

At = 0.5 m/s^2 ( At = Tangential acceleration )

Speed of the cyclist,v=27 km/hr = 275/18 = 7.5 m/s

Radius of the circular turn, r=80m

Centripetal acceleration (Ac) is given as:

Ac = v^2/r

Ac = ((7.5)^2)/80 = 0.7 m/s^2

Since the angle between Ac and At is 90degrees

Therefore, the resultant or net acceleration (A) is given by :-

A = (Ac^2 + At^2)^0.5

A = (0.7^2+0.5^2)^ 0.5 = 0.86 m/s^2

Related Questions

Amity University, Noida Law A...
Apply
700+ Campus placements at top national and global law firms, corporates and judiciaries
Amity University, Noida BBA A...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
VIT Bhopal University | M.Tec...
Apply
M.Tech admissions open @ VIT Bhopal University | Highest CTC 52 LPA | Apply now
Amity University | M.Tech Adm...
Apply
Ranked amongst top 3% universities globally (QS Rankings).
Graphic Era (Deemed to be Uni...
Apply
NAAC A+ Grade | Among top 100 universities of India (NIRF 2024) | 40 crore+ scholarships distributed
Amity University Noida B.Tech...
Apply
Among Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books