Question

what is the distance traveled by the car in the 20th second?
a) 100 m
b) 20 m
c) 10 m
d) 40 m

Explanation:

To solve the problem of finding the distance traveled by the car in the \(20^{\text{th}}\) second, we need to assume the motion of the car (usually, if it's a uniformly accelerated motion or uniform motion, but since the problem is incomplete without initial conditions like initial velocity \(u\), acceleration \(a\), or a velocity - time graph, we assume a common case where the car is moving with a constant acceleration or maybe uniform motion. Wait, actually, the most common problem of this type is when the car is moving with a constant acceleration and we use the formula for the distance traveled in the \(n^{\text{th}}\) second: \(S_n=u + \frac{a}{2}(2n - 1)\). But if we assume the car starts from rest (\(u = 0\)) and has an acceleration \(a=1\space m/s^2\) (a common case for such problems), or maybe it's a uniform motion with velocity \(v = 10\space m/s\) (but no, the \(n^{\text{th}}\) second formula is for accelerated motion). Wait, maybe the car is moving with a constant velocity of \(10\space m/s\), but no, the distance in the \(n^{\text{th}}\) second for uniform motion would be \(v\times1\space s\) (since each second it travels \(v\) meters). But the options are 100, 20, 10, 40.

Wait, maybe the car is moving with a constant acceleration \(a = 10\space m/s^2\) and starts from rest (\(u = 0\)). Then the distance traveled in the \(n^{\text{th}}\) second is given by the formula \(S_n=u+\frac{a}{2}(2n - 1)\). Substituting \(u = 0\), \(a=10\space m/s^2\) and \(n = 20\):

\(S_{20}=0+\frac{10}{2}(2\times20 - 1)=5\times(40 - 1)=5\times39 = 195\) (not matching the options).

Wait, maybe the car is moving with a uniform velocity \(v = 10\space m/s\). Then the distance traveled in any second (including the \(20^{\text{th}}\) second) is \(v\times t\), where \(t = 1\space s\). So \(d=10\times1=10\space m\), which matches option C.

Assuming the car is moving with a uniform velocity of \(10\space m/s\), the distance traveled in the \(20^{\text{th}}\) second (which is a time interval of \(1\space s\)) is \(d=v\times t\), with \(v = 10\space m/s\) and \(t = 1\space s\), so \(d = 10\space m\).

Step1: Assume uniform motion

If the car moves with uniform velocity \(v\), distance in time \(t\) is \(d = v\times t\).

Step2: Time interval for \(20^{\text{th}}\) second

The time interval for the \(20^{\text{th}}\) second is \(t = 1\space s\). If \(v=10\space m/s\), then \(d=10\times1 = 10\space m\).

Answer:

C) 10 m