×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

The position of a particle as a function of time is given by x(t)=ln(2t+3) for all positive t , where x is in meters and t is in seconds. What is the particle's acceleration at t=10sec ? A −231​m/s2 B −232​m/s2 C −5294​m/s2 D −234​m/s2 E −51​m/s2


3 months ago

ankit singh
600 Points
							ANSWERPosition of the particle            x(t)=ln(2t+3)  mDifferentiating w.r.t time we get its velocity        v(t)=dtd[x(t)]​=2t+32​    m/s Differentiating again w.r.t time we get its acceleration        a(t)=dtd[v(t)]​=2×(2t+3)2−2​=(2t+3)2−4​    m/s2⟹   a∣∣∣∣∣​t=10s​=(2×10+3)2−4​=529−4​   m/

3 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question