Question

Approximately, how long would it take a car, starting from rest and accelerating uniformly in a straight line at 5
m/s2, to cover a distance of 200 m.?​

Answer 1

Proper QuEstion: Approximately, how long would it take a car, starting from rest and accelerating uniformly in a straight line at 5 m/s², to cover a distance of 200 m?

Understanding the question: This question says that we have to calculate the time in which a car, starting from rest and accelerating uniformly in a straight line at 5 m/s², to cover a distance of 200 metres. Let's solve this question!

ProvidEd that:

• Initial velocity = 0 m/s
• Acceleration = 5 m/s sq.
• Distance = 200 metres

Don't be confused! Initial velocity cames as zero because the car starts from rest.

To calculaTe:

• The time taken

SoluTion:

• The time taken = 8.94 seconds

UsiNg concept:

• Second equation of motion

UsiNg formula:

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}$

(Where, s denotes displacement or distance or height, u denotes initial velocity, t denotes time taken, a denotes acceleration)

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf 200 \: = 0(t) + \dfrac{1}{2} \times 5(t)^{2} \\ \\ :\implies \sf 200 \: = 0 + \dfrac{1}{2} \times 5(t)^{2} \\ \\ :\implies \sf 200 \: = 0 + \dfrac{1}{\cancel{{2}}} \times \cancel{5}(t)^{2} \\ \\ :\implies \sf 200 \: = 0 + 1 \times 2.5(t)^{2} \\ \\ :\implies \sf 200 \: = 0 + 2.5(t)^{2} \\ \\ :\implies \sf 200 \: = 2.5(t)^{2} \\ \\ :\implies \sf \dfrac{200}{2.5} \: = (t)^{2} \\ \\ :\implies \sf \cancel{\dfrac{200}{2.5}} \: = (t)^{2} \\ \\ :\implies \sf 80 \: = (t)^{2} \\ \\ :\implies \sf \sqrt{80} \: = t \\ \\ :\implies \sf 8.94 \: = t \\ \\ :\implies \sf t \: = 8.94 \: seconds \\ \\ :\implies \sf Time \: = 8.94 \: seconds$