Skip to content
Numerical · Q27

Q.A car of mass 1500 kg1500\ \text{kg} negotiates a curve of radius 250 m250\ \text{m}, banked at 10∘10^\circ, where the coefficient of static friction between the tyres and the road is 0.30.3. Find the maximum speed at which the car can travel around the curve without skidding. (Take g=9.8 m/s2g = 9.8\ \text{m/s}^2.)

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
84% · 27/32 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Given: r=250 mr = 250\ \text{m}, θ=10∘\theta = 10^\circ, μs=0.3\mu_s = 0.3, g=9.8 m/s2g = 9.8\ \text{m/s}^2 (the car's mass, 1500 kg1500\ \text{kg}, cancels out of the result and is not needed).\n\nUsing the banked-road-with-friction maximum-speed formula, vmax⁡=rg(tan⁡θ+μs)1−μstan⁡θv_{\max} = \sqrt{\frac{rg(\tan\theta + \mu_s)}{1 - \mu_s\tan\theta}} with tan⁡10∘≈0.1763\tan 10^\circ \approx 0.1763:\n\nNumerator: rg(tan⁡θ+μs)=250×9.8×(0.1763+0.3)=2450×0.4763≈1167.0rg(\tan\theta+\mu_s) = 250 \times 9.8 \times (0.1763+0.3) = 2450 \times 0.4763 \approx 1167.0\n\nDenominator: 1−μstan⁡θ=1−0.3×0.1763≈1−0.0529=0.94711 - \mu_s\tan\theta = 1 - 0.3 \times 0.1763 \approx 1 - 0.0529 = 0.9471\n\n$$v_{\max}^2 = \frac{1167.0}{0.9471} \approx 1232 …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.